CalcIII SurfaceIntegrals Solutions

CALCULUS III
Solutions to Practice Problems

Surface Integrals
Paul Dawkins
Calculus III
Table of Contents
Preface ............................................................................................................................................ ii
Surface Integrals............................................................................................................................ 3
Parametric Surfaces.................................................................................................................................... 3
Surface Integrals .......................................................................................................................................14
Surface Integrals of Vector Fields .............................................................................................................35
Stokes’ Theorem .......................................................................................................................................61
Divergence Theorem .................................................................................................................................77
© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

Calculus III
Preface
Here are the solutions to the practice problems for my Calculus II notes. Some solutions will
have more or less detail than other solutions. As the difficulty level of the problems increases
less detail will go into the basics of the solution under the assumption that if you’ve reached the
level of working the harder problems then you will probably already understand the basics fairly
well and won’t need all the explanation.
This document was written with presentation on the web in mind. On the web most solutions are
broken down into steps and many of the steps have hints. Each hint on the web is given as a
popup however in this document they are listed prior to each step. Also, on the web each step can
be viewed individually by clicking on links while in this document they are all showing. Also,
there are liable to be some formatting parts in this document intended for help in generating the
web pages that haven’t been removed here. These issues may make the solutions a little difficult
to follow at times, but they should still be readable.
© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

Calculus II
Surface Integrals
Parametric Surfaces
1. Write down a set of parametric equations for the plane 7 x + 3 y + 4 z =

15 .
Step 1
There isn’t a whole lot to this problem. There are three different acceptable answers here. To get a set of
parametric equations for this plane all we need to do is solve for one of the variables and then write down
the parametric equations.
For this problem let’s solve for z to get,
z =154 − 74 x − 43 y
Step 2
The parametric equation for the plane is then,

r ( x,=
y) x, y=
,z x, y, 154 − 74 x − 43 y
Remember that all we need to do to get the parametric equations is plug in the equation for z into the z
component of the vector x, y, z .
Also, as noted in Step 1 we could just have easily done either of the following two forms for the
parametric equations for this plane.
 
=r ( x, z ) x, g ( x, z ) , z
= r ( y, z ) h ( y, z ) , y, z
where you solve the equation of the plane for y or x respectively. All three set of parametric equations are
all perfectly valid forms for the answer to this problem.
2. Write down a set of parametric equations for the plane 7 x + 3 y + 4 z =

15 that lies in the 1st octant.
Step 1
This problem is really just an extension of the previous problem so we’ll redo the set of parametric
equations for the plane a little quicker this time.
First we need to solve the equation for any of the three variables. We’ll solve for z in this case to get,
© 2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx

Calculus II
z =154 − 74 x − 43 y
The parametric equation for this plane is then,


r ( x, =
y) x, y,=
z x, y, 154 − 74 x − 43 y
Remember that all we need to do to get the parametric equations is plug in the equation for z into the z
Step 2
Now, the set of parametric equations from above is for the full plane and that isn’t what we want
in this problem. In this problem we only want the portion of the plane that is in the 1st octant.
So, we’ll need to restrict x and y so that the parametric equation from Step 1 will only give the portion of
the plane that is in the 1st octant.
If you recall how to get the region D for a triple integral then you know how to do this because it is
basically the same idea. In this case we need the region D in the xy-plane that will give the plane in the 1st
octant.
Here is a sketch of this region.
The hypotenuse is just where the plane intersects the xy-plane and so we can quickly find the equation of
the line by setting z = 0 in the equation of the plane.
We can either solve this for x or y to get the ranges for x and y. It doesn’t really matter which we solve
for here so let’s just solve for y to get the following ranges for x and y to describe this triangle.
0 ≤ x ≤ 157
0 ≤ y ≤ − 73 x + 5
Putting this all together we get the following set of parametric equations for the plane that is in the 1st
octant.

Calculus II

r=( x, y ) x, y, 154 − 74 x − 34 y 0 ≤ x ≤ 157 , 0 ≤ y ≤ − 73 x + 5
5 for −1 ≤ z ≤ 6 .
3. The cylinder x 2 + y 2 =
Step 1
Because this surface is just a cylinder we just need the cylindrical coordinates conversion formulas with
the polar coordinates in the xy-plane (since the cylinder is given in terms of x and y).
The conversion equations are,
= cos θ
x r= sin θ
y r= z z
However, recall that we are actually on the surface of the cylinder and so we know that r = 5 . The
conversion equations are then,
=x 5 cos θ
= y 5 sin θ
= z z
Step 2
We can now write down a set of parametric equations for the cylinder. They are,

r= ( z,θ ) =
x, y , z 5 cos θ , 5 sin θ , z
Remember that all we do is plug the conversion formulas for x, y, and z into the x, y and z components of
the vector x, y, z and we have a set of parametric equations. Also note that because the resulting
vector equation is an equation in terms of z and θ those will also be the variables for our set of
parametric equation.
Step 3
Now, the only issue with the set of parametric equations above is that they are for the full cylinder and we
don’t want that. We only want the cylinder in the given range of z so to finish this problem out all we
need to do is add on a set of restrictions or ranges to our variables.
Doing that gives,

=r ( z,θ ) 5 cos θ , 5 sin θ , z − 1 ≤ z ≤ 6 , 0 ≤ θ ≤ 2π
Note that the z range is just the range given in the problem statement and the θ range is the full zero to
2π range since there was no mention of restricting the portion of the cylinder that we wanted with
respect to θ (for example, only the top half of the cylinder).

Calculus II
4. The portion of y =4 − x 2 − z 2 that is in front of y = −6 .
Step 1
Okay, the basic set of parametric equations in this case is pretty easy since we already have the equation
in the form of “y = ”.
The set of parametric equations that will give the full surface is just,

r ( x, z )= x, y , z = x, 4 − x 2 − z 2 , z
Remember that all we need to do to get the parametric equations is plug in the equation for y into the y
Step 2
Finally, all we need to do is restrict x and z to get only the portion of the surface we are looking for. That
is pretty simple however since we are given that we only want the portion that is in front of y = −6 .
This is equivalent to requiring that y ≥ −6 and we do have the equation of the surface so all we need to
do is plug that into the inequality and do a little rewrite. Doing this gives,
4 − x 2 − z 2 ≥ −6 → x 2 + z 2 ≤ 10
In other words we only want the points ( x, z ) that are inside the disk of radius 10 .
Putting all of this together gives the following set of parametric equations for the portion of the surface
we are after.

r ( x, z )= x, 4 − x 2 − z 2 , z x 2 + z 2 ≤ 10
5. The portion of the sphere of radius 6 with x ≥ 0 .
Step 1
Because we have a portion of a sphere we’ll start off with the spherical coordinates conversion formulas.
x ρ=
sin ϕ cos θ y ρ=
sin ϕ sin θ z ρ cos ϕ
However, we are actually on the surface of the sphere and so we know that ρ = 6 . With this the
conversion formulas become,
x =
6sin ϕ cos θ =
y 6sin ϕ sin θ z 6 cos ϕ
Step 2
The set of parametric equations that will give the full sphere is then,

Calculus II

r=(θ , ϕ ) =
x, y , z 6sin ϕ cos θ , 6sin ϕ sin θ , 6 cos ϕ
Remember that all we do is plug the conversion formulas for x, y, and z into the x, y and z components of
the vector x, y, z and we have a set of parametric equations. Also note that because the resulting
vector equation is an equation in terms of θ and ϕ those will also be the variables for our set of
parametric equation.
Step 3
Finally, we need to deal with the fact that we don’t actually want the full sphere here. We only want the
portion of the sphere for which x ≥ 0 .
We can restrict x to this range if we restrict θ to the range − 12 π ≤ θ ≤ 12 π .
We’ve not put any restrictions on z and so that means that we’ll take the full range of possible ϕ or
0 ≤ ϕ ≤ π . Recall that ϕ is the angle a point in spherical coordinates makes with the positive z-axis and
so that is the quantity we’d need to restrict if we’d wanted to restrict z (for example z ≤ 0 ).
Putting all of this together gives the following set of parametric equations for the portion of the surface
we are after.

r (θ , ϕ ) 6sin ϕ cos θ , 6sin ϕ sin θ , 6 cos ϕ − 12 π ≤ θ ≤ 12 π , 0 ≤ ϕ ≤ π
6. The tangent plane to the surface given by the following parametric equation at the point ( 8,14, 2 ) .
   
r ( u , v ) = ( u 2 + 2u ) i + ( 3v − 2u ) j + ( 6v − 10 ) k
Step 1
In order to write down the equation of a plane we need a point, which we have, ( 8,14, 2 ) , and a normal
vector, which we don’t have yet.
 
However, recall that ru × rv will be normal to the surface. So, let’s compute that.
     
ru =( 2u + 2 ) i − 2 j rv =3 j + 6k
  
i j k
    
ru × rv =2u + 2 −2 0 =−12i − 6 ( 2u + 2 ) j + 3 ( 2u + 2 ) k
0 3 6
Step 2

Calculus II
 
Now having ru × rv is all well and good but it is really only useful if we also know the point, ( u , v ) for
which we are at ( 8,14, 2 ) so we next need to set the x, y and z coordinates of our point equal to the x, y
and z components of our parametric equation to determine the value of u and v we need.
Here are the equations we get if we do that.
8 = u 2 + 2u 0 = u 2 + 2u − 8 = ( u + 4 )( u − 2 )
14 = 3v − 2u ⇒ 14 = 3v − 2u
2= 6v − 10 12 = 6v
Step 3
From the third equation above we can see that we must have v = 2 and from the first equation we can see
that we must have either u = −4 or u = 2 .
Plugging our only choice for v and both choices for u into the second equation we can see that we must
have u = −4 .
Step 4
 
Plugging u = −4 and v = 2 into the equation for ru × rv we will arrive at the following normal vector to
the surface at ( 8,14, 2 ) .
     
( ru rv )u =
n =× −4, v =
2
=
−12 i + 36 j − 18 k
Note that, in this case, the normal vector didn’t actually depend on the value of v. That won’t happen in
general, but as we’ve seen here that kind of thing can happen on occasion so don’t get excited about it
when it does.
   
The equation of the tangent plane to the surface at ( 8,14, 2 ) with normal vector n =
−12i + 36 j − 18k
is,
−12 ( x − 8 ) + 36 ( y − 14 ) − 18 ( z − 2 )= 0 → − 12 x + 36 y − 18 z= 372
Step 5
To get a set of parametric equations for the tangent plane all we need to do is solve the equation for z to
get,
z=− 623 + 23 x − 2 y
We can then plug this into the vector x, y, z to get the following set of parametric equations for the
tangent plane.

r ( x, y=
) x, y, − 623 + 32 x − 2 y

Calculus II
Note that there will be no restrictions on x and y because we wanted the full tangent plane.
9 that is inside the cylinder x 2 + y 2 =

7. Determine the surface area of the portion of 2 x + 3 y + 6 z = 7.
Step 1
We first need to parameterize the surface. Because we are wanting the portion that is inside the cylinder
centered on the z-axis it makes sense to first solve the equation of the plane for z to get,
z =32 − 13 x − 12 y
The parameterization for the full plane is then,


y)
r ( x, = x, y, 32 − 13 x − 12 y
We only want the portion that is inside the cylinder given in the problem statement so we’ll also need to
restrict x and y to those in the disk x 2 + y 2 ≤ 7 . This will now give only the portion of the plane that is
inside the cylinder.
Step 2
 
Next we need to compute rx × ry . Here is that work.
 
rx =
1, 0, − 13 ry =
0,1, − 12
  
i j k
    
rx × ry = 1 0 − 13 = 1
3 i + 12 j + k
0 1 − 12
Now, we what we really need is,
 
( 13 ) + ( 12 )
2 2
rx × ry= + 1= =
49
36
7
6
Step 3
The integral for the surface area is then,
A = ∫∫ 76 dA
D
In this case D is just the restriction on x and y that we noted in Step 1. So, D is just the disk x 2 + y 2 ≤ 7 .
Step 4
Computing the integral in this case is very simple. All we need to do is take advantage of the
fact that,

Calculus II
∫∫ dA = Area of D
D
So, the surface area is simply,
∫∫ dA [ Area =
of D ] ( )
π =
7 
2
=A ∫∫=
D
dA =
7
6
7
6
D
7
6
7
6
 
49
6 π
25 with z ≤ 0 .
8. Determine the surface area of the portion of x 2 + y 2 + z 2 =
Step 1
We first need to parameterize the sphere and we’ve already done a sphere in this problem set so we won’t
go into great detail with the parameterization here.
The parameterization for the full sphere is,


r (θ , ϕ ) = 5sin ϕ cos θ ,5sin ϕ sin θ ,5cos ϕ
We don’t want the full sphere of course. We only want the lower half of the sphere, i.e. the portion with
z ≤ 0 . This means that we’ll need to restrict ϕ to 12 π ≤ ϕ ≤ π . Recall that ϕ is the angle points make
with the positive z-axis and because we only want points below the xy-plane we’ll need the range of
2π ≤ϕ ≤π
1
.
We want the full lower half and so we’ll use 0 ≤ θ ≤ 2π for our θ range.
Step 2
 
Next we need to compute rθ × rϕ . Here is that work.
 
−5sin ϕ sin θ ,5sin ϕ cos θ , 0
rθ = 5cos ϕ cos θ ,5cos ϕ sin θ , −5sin ϕ
rϕ =
  
i j k
 
rθ × rϕ =−5sin ϕ sin θ 5sin ϕ cos θ 0
5cos ϕ cos θ 5cos ϕ sin θ −5sin ϕ
   
−25sin 2 ϕ cos θ i − 25sin ϕ cos ϕ sin 2 θ k − 25sin ϕ cos ϕ cos 2 θ k − 25sin 2 ϕ sin θ j
=
  
−25sin 2 ϕ cos θ i − 25sin 2 ϕ sin θ j − 25sin ϕ cos ϕ ( sin 2 θ + cos 2 θ ) k
=
  
−25sin 2 ϕ cos θ i − 25sin 2 ϕ sin θ j − 25sin ϕ cos ϕ k
=

Calculus II
 
( −25sin ϕ cos θ ) + ( −25sin 2 ϕ sin θ ) + ( −25sin ϕ cos ϕ )
2 2
rθ × rϕ = 2 2
= 625sin 4 ϕ ( cos 2 θ + sin 2 θ ) + 625sin 2 ϕ cos 2 ϕ
= 625sin 2 ϕ ( sin 2 ϕ + cos 2 ϕ )

= 25 sin ϕ
= 25sin ϕ
Note that we can drop the absolute value bars on the sine because we know that sine will be positive in
2π ≤ϕ ≤π
1
.
Step 3
2π π
=A 25sin ϕ dA ⌠
∫∫D=  ∫ π 25sin ϕ dϕ dθ
⌡0 1
2
As noted in the integral above D is just the ranges of θ and ϕ we found in Step 1.
Step 4
Now we just need to evaluate the integral to get the surface area.
π 2π 2π 2π
⌠
 ∫1 25sin ϕ dϕ dθ =
π
A=
⌡0 2 π ∫0 2
∫
−25cos ϕ 1 π dθ ==
0
25 dθ 50π
1 4
9. Determine the surface area of the portion of z =3 + 2 y + x that is above the region in the xy-plane
4
bounded by y = x 5 , x = 1 and the y-axis.
Step 1
Parameterizing this surface is pretty simple. We have the equation of the surface in the form
z = f ( x, y ) and so the parameterization of the surface is,

r ( x, y=
) x, y,3 + 2 y + 14 x 4
Now, this is the parameterization of the full surface and we only want the portion that lies over the
following region.

Calculus II
So, to get only the portion of the surface we’ll need to restrict x and y to the following ranges,
0 ≤ x ≤1
0 ≤ y ≤ x5
On a side note we can see that we are in the 1st quadrant here and so we know that x ≥ 0 and y ≥ 0 .
Therefore we can see that the surface in the 1st quadrant is always above the xy-plane and so will in fact
always be above the region above as suggested in the problem statement.
Step 2
 
Next we need to compute rx × ry . Here is that work.
 
=rx =
1, 0, x3 ry 0,1, 2
  
i j k
    
rx × ry =1 0 x3 =− x 3i − 2 j + k
0 1 2
 
rx × ry = ( − x ) + ( −2 ) + (1)
3 2 2 2
= x6 + 5
Step 3
1 5
x 6 + 5 dA= ⌠ ∫
x
A= ∫∫
D
⌡0 0
x 6 + 5 dy dx
As noted in the integral above D is just the ranges of x and y we found in Step 1.
Step 4

Calculus II
1 5
x + 5 dy dx = ⌠
1 5 x
A= ⌠ ∫
x
 y x + 5 dx
6 6
⌡0 0 ⌡0 0
(6 )
3 1
(x + 5)
1
∫
3 3
= x 5
x + 5 dx =
6 1
9
6 2
= 1
9
2
− 5 2 = 0.3907
0
0
10. Determine the surface area of the portion of the surface given by the following parametric equation
that lies inside the cylinder u 2 + v 2 =
4.

r ( u, v )
= 2u , vu ,1 − 2v
Step 1
We’ve already been given the parameterization of the surface in the problem statement so we don’t need
to worry about that for this problem. All we really need to do yet is to acknowledge that we’ll need to
restrict u and v to the disk u 2 + v 2 ≤ 4 .
Step 2
 
Next we need to compute ru × rv . Here is that work.
 
=
ru 2, v, 0 =
rv 0, u , −2
  
i j k
    
ru × rv =2 v 0 =−2vi + 4 j + 2uk
0 u −2
 
ru × rv= ( −2v ) + ( 4 ) + ( 2u )= 4u 2 + 4v 2 + 16= 2 u 2 + v 2 + 4
2 2 2
Step 3
=A ∫∫ 2
D
u 2 + v 2 + 4 dA
Where D is the disk u 2 + v 2 ≤ 4 .
Step 4
Because D is a disk the best bet for this integral is to use the following “version” of polar coordinates.
=u r cos θ =v r sin θ u 2= dA r dr dθ
+ v2 r 2 =

Calculus II
The polar coordinate limits for this D is,
0 ≤ θ ≤ 2π
0≤r ≤2
So, the integral to converting to polar coordinates gives,
2π 2
= ∫∫ 2 u + v + 4= ∫ ∫ 2r r 2 + 4 dr dθ
2 2
A dA
0 0
D
Step 5
2π
⌠ 3 2
(r + 4)
2π 2
A= ∫ ∫ 2r r + 4 dr dθ =  dθ
2 2 2 2
⌡0
0 0 3
0
( ) ( )
2π
( )
2π
= ⌠
3 3 3
2
8 2 − 4 2 dθ= 2
82 − 8 θ = 32
8 − 1= 61.2712
⌡0 3 3
0
3
Surface Integrals
∫∫ z + 3 y − x dS where S is the portion of z =2 − 3 y + x 2 that lies over the triangle in the

2
1. Evaluate
S
xy-plane with vertices ( 0, 0 ) , ( 2, 0 ) and ( 2, −4 ) .
Step 1
Let’s start off with a quick sketch of the surface we are working with in this problem.

Calculus II
We included a sketch with traditional axes and a sketch with a set of “box” axes to help visualize the
surface.
The orange surface is the sketch of z =2 − 3 y + x 2 that we are working with in this problem. The
greenish triangle below the surface is the triangle referenced in the problem statement that lies below the
surface. This triangle will be the region D for this problem.
Here is a quick sketch of D just to get a better view of it than the mostly obscured view in the sketch
above.

Calculus II
We could use either of the following sets of limits to describe D.
0≤ x≤2 −4≤ y ≤0
−2 x ≤ y ≤ 0 − 12 y ≤ x ≤ 2
We’ll decide which set to use in the integral once we get that set up.
Step 2
Let’s get the integral set up now. In this case the surface is in the form,
z =g ( x, y ) =2 − 3 y + x 2
so we’ll use the following formula for the surface integral.
⌠⌠  ∂g   ∂g 
2 2
 (
∫∫S f ( x, y, z ) dS 
= f x, y, g ( x, y ) )   +   + 1 dA
⌡⌡  ∂x   ∂y 
D
The integral is then,
∫∫ z + 3 y − x = ⌠⌠ ( 2 − 3 y + x 2 ) + 3 y − x 2  ( 2 x ) + ( −3) + 1 dA
2 2 2
dS
⌡⌡
S D
= ∫∫ 2
D
4 x 2 + 10 dA
Don’t forget to plug the equation of the surface into z in the integrand and recall that D is the triangle
sketched in Step 1.
Step 3
Now all that we need to do is evaluate the double integral and that shouldn’t be too difficult at this point.
First note that from the integrand it should be pretty clear that we’ll want to integrate with respect to y
first (unless you want to do a trig substitution of course….). So, the integral becomes,

Calculus II
∫∫ z + 3 y − x=
dS ∫∫ 2 4 x 2 + 10 dA
2
S D
2 0
= ∫∫ 0 −2 x
2 4 x 2 + 10 dy dx
( )
2
= ⌠
0
 2 y 4 x + 10
2
dx
⌡0 −2 x
2
= ∫ 0
4 x 4 x 2 + 10 dx
( 26 )
3 2
( 4 x 2 + 10 ) 2 =
3 3
= 1
3
1
3
2
− 10 2 = 33.6506
0
2. Evaluate ∫∫ 40 y dS
S
where S is the portion of=
y 3 x 2 + 3 z 2 that lies behind y = 6 .
Step 1

Calculus II
Note that the surface in this problem is only the elliptic paraboloid and does not include the “cap” at
y = 6 . We would only include the “cap” if the problem had specified that in some manner to make it
clear.
In this case D will be the circle/disk we get by setting the two equations equal or,
6 = 3x 2 + 3z 2 ⇒ x2 + z 2 = 2
So, D will be the disk x 2 + z 2 ≤ 2 .
Step 2
Let’s get the integral set up now. In this case the surface is in the form,
y g ( x,=
= z ) 3x 2 + 3z 2
so we’ll use the following formula for the surface integral.
⌠⌠  ∂g 
2
 ∂g 
2
∫∫S f ( x, y, z ) dS  f ( x, g ( x, z ) , z )  ∂x  + 1 +  ∂z  dA
=
⌡⌡
D

Calculus II
40 y dS ∫∫ 40 ( 3 x
∫∫ = + 3z 2 ) (6x) + 1 + ( 6 z ) dA
2 2 2
S D
= ∫∫ 120 ( x + z 2 ) 36 ( x 2 + z 2 ) + 1 dA
2
Don’t forget to plug the equation of the surface into y in the integrand and recall that D is the disk we
found in Step 1.
Step 3
Now, for this problem it should be pretty clear that we’ll want to use polar coordinates to do the integral.
We’ll use the following set of polar coordinates.
= θ
x r cos= z r sin θ =x2 + z 2 r 2
Also, because D is the disk x 2 + z 2 ≤ 2 the limits for the integral will be,
0 ≤ θ ≤ 2π
0≤r ≤ 2
Converting the integral to polar coordinates gives,
2π
∫∫S 40 y dS ⌠⌡0 120r 2 36r 2 + 1 ( r ) dr dθ

2
= ∫
0
2π
= ⌠
2
⌡0 ∫
0
120r 3 36r 2 + 1 dr dθ
Don’t forget to pick up the extra r when converting the dA into polar coordinates.
Step 4
Now all that we need to do is evaluate the double integral and this one can be a little tricky unless you’ve
seen this kind of integral done before.
We’ll use the following substitution to do the integral.
u =36r 2 + 1 → du =72r dr → 1
72 du =r dr
The problem is that this doesn’t seem to work at first glance because the differential will only get rid of
one of the three r’s in front of the root. However, we can also solve the substitution for r 2 to get,
=
r2 1
36 ( u − 1)
and we can now convert the remaining two r’s into u’s.
So, using the substitution the integral becomes,

Calculus II
2π
120 ( 721 )( 361 ) ( u − 1) u 2 du dθ
73
∫∫ 40 y dS ∫ ∫
1
=
0 1
S
(u )
2π
= ⌠
73
 ⌠
3 1
5 2
− u 2 du dθ
⌡0 ⌡1 108
Note that we also converted the r limits in the original integral into u limits simply by plugging the “old”
r limits into the substitution to get “new” u limits.
We can now easily finish evaluating the integral.
(
) dθ
2π
⌠
73
∫∫S 40 y dS ⌡0
5 3
= 5
108
2
5 u 2 − 32 u 2
1
 ( ) ( )
2π
= ⌠
  73 − 73 − ( − )  dθ
5 2
5
2 2
3
2 4
⌡ 0
108 5
 3 15
=
 (
5π
54
2
5) ( ) +  5176.8958
 73 − 73 = 5
2 2
3
3
2 4
15
Kind of messy integral with a messy answer but that will happen on occasion so we shouldn’t get too
excited about it when that does happen.
3. Evaluate ∫∫ 2 y dS
S
4 between x = 0 and x= 3 − z .
where S is the portion of y 2 + z 2 =
Step 1

Calculus II
Note that the surface in this problem is only the cylinder itself. The “caps” of the cylinder are not part of
this surface despite the red “cap” in the sketch. That was included in the sketch to make the front edge of
the cylinder clear in the sketch. We would only include the “caps” if the problem had specified that in
some manner to make it clear.
Step 2
Now because our surface is a cylinder we’ll need to parameterize it and use the following
formula for the surface integral.
  
f ( x, y, z ) dS ∫∫ f ( r ( u , v ) ) r × r
∫∫= u v dA
S D
where u and v will be chosen as needed when doing the parameterization.
We saw how to parameterize a cylinder in the previous section so we won’t go into detail for the
parameterization. The parameterization is,

r ( x, θ ) = x, 2sin θ , 2 cos θ 0 ≤ θ ≤ 2π , 0 ≤ x ≤ 3 − z = 3 − 2 cos θ
We’ll use the full range of θ since we are allowing it to rotate all the way around the x-axis. The x limits
come from the two planes that “bound” the cylinder and we’ll need to convert the upper limit using the
parameterization.
Next, we’ll need to compute the cross product.

 
=rx 1, 0, 0 =rθ 0, 2 cos θ , −2sin θ

Calculus II
  
i j k
   
rx =
× rθ 1 0 0= 2sin θ j + 2 cos θ k
0 2 cos θ −2sin θ
The magnitude of the cross product is,
 
rx =
× rθ 4sin 2 θ + 4 cos
= 2
θ 2
=∫∫ 2 y dS 2 ( 2sin θ )( 2 ) dA ∫∫ 8sin θ dA

∫∫=
S D D
Don’t forget to plug the y component of the surface parametrization into the integrand and D is just the
limits on x and θ we noted above in the parameterization.
Step 3
2π 3− 2cosθ
∫∫ 2 y dS = ∫ ∫
S
0 0
8sin θ dx dθ
2π 3− 2cosθ
=∫
0
(8 x sin θ ) 0 dθ
2π
∫ 8 ( 3 − 2 cos θ ) sin θ dθ
=
0
π 2
= ∫ 24sin θ − 16sin θ cos θ dθ
0
π
= ∫ 24sin θ − 8sin ( 2θ ) dθ
2
( −24 cos θ + 4 cos ( 2θ ) ) =

π 2
= 0
0
4. Evaluate ∫∫ xz dS
S
where S is the portion of the sphere of radius 3 with x ≤ 0 , y ≥ 0 and z ≥ 0 .
Step 1

Calculus II
Note that the surface in this problem is only the part of the sphere itself. The “edges” (the greenish
portions on the right/left) are not part of this surface despite the fact that they are in the sketch. They

Calculus II
were included in the sketch to try and make the surface a little clearer in the sketch. We would only
include the “edges” if the problem had specified that in some manner to make it clear.
Step 2
Now because our surface is a sphere we’ll need to parameterize it and use the following formula
for the surface integral.
  
f ( x, y, z ) dS ∫∫ f ( r ( u , v ) ) r × r
∫∫= u v dA
S D
where u and v will be chosen as needed when doing the parameterization.
We saw how to parameterize a sphere in the previous section so we won’t go into detail for the

r (θ , ϕ ) 3sin ϕ cos θ ,3sin ϕ sin θ ,3cos ϕ 1
2 π ≤ θ ≤ π , 0 ≤ ϕ ≤ 12 π
We needed the restriction on ϕ to make sure that we only get a portion of the upper half of the sphere
(i.e. z ≥ 0 ). Likewise the restriction on θ was needed to get only the portion that was in the 2nd quadrant
of the xy-plane (i.e. x ≤ 0 and y ≥ 0 ).

 
rϕ =
  
i j k
 
   
−9 sin 2 ϕ cos θ i − 9 sin ϕ cos ϕ sin 2 θ k − 9 sin ϕ cos ϕ cos 2 θ k − 9 sin 2 ϕ sin θ j
=
  
=
  
=
The magnitude of the cross product is,
 
( −9sin ϕ cos θ ) + ( −9sin 2 ϕ sin θ ) + ( −9sin ϕ cos ϕ )
2 2
rθ × rϕ = 2 2
= 81sin 4 ϕ ( cos 2 θ + sin 2 θ ) + 81sin 2 ϕ cos 2 ϕ
= 81sin 2 ϕ ( sin 2 ϕ + cos 2 ϕ )

= 9 sin ϕ
= 9sin ϕ

Calculus II
( 3sin ϕ cos θ )( 3cos ϕ )( 9sin ϕ ) dA ∫∫ 81cos ϕ sin

∫∫ xz dS ∫∫=
2
ϕ cos θ dA
S D D
Don’t forget to plug the x and z component of the surface parameterization into the integrand and D is just
the limits on θ and ϕ we noted above in the parameterization.
Step 3
∫∫ xz dS = ∫∫ 81cos ϕ sin ϕ cos θ dA

2
S D
π 1π
= ⌠ ∫ 81cos ϕ sin 2 ϕ cos θ dϕ dθ

2
⌡ 12 π 0
π
= ⌠ ( 27 sin 3 ϕ cos θ dϕ ) dθ
1π
2
⌡ 12 π 0
π
= ∫1 27 cos θ dθ
2
π
π
= ( 27 sin θ ) 1π
2
= −27
5. Evaluate ∫∫ yz + 4 xy dS
S
where S is the surface of the solid bounded by 4 x + 2 y + z =8 , z = 0,
y = 0 and x = 0 . Note that all four surfaces of this solid are included in S.
Step 1

Calculus II
Okay, as noted in the problem statement all four surfaces in the sketch (two not shown) are part of S so
let’s define each of them as follows.
S1 : Plane given by 4 x + 2 y + z =8 ( i.e the top of the solid )

S 2 : Plane given by y = 0 ( i.e the triangle on right side of the solid )
S3 : Plane given by x = 0 ( i.e the triangle at back of the solid - not shown in sketch )
S 4 : Plane given by z = 0 ( i.e the triangle on bottom of the solid - not shown in sketch )
As noted in the definitions above the first two surfaces are shown in the sketch but the last two are not
actually shown due to the orientation of the solid. Below are sketches of each of the three surfaces that
correspond to the coordinates planes.

Calculus II
With each of the sketches we gave limits on the variables for each of them since we’ll eventually need
that when we start doing the surface integral along each surface.
Now we need to go through and do the integral for each of these surfaces and we’re going to go through
these a little quicker than we did for the first few problems in this section.
Step 2
Let’s start with S1 . In this case the surface can easily be solved for z to get,
z =8 − 4 x − 2 y
With the equation of the surface written in this manner the region D will be in the xy-plane and if you
think about it you’ll see that in fact D is nothing more than S 4 !
The integral in this case is,

Calculus II
= ∫∫  y ( 8 − 4 x − 2 y ) + 4 xy  ( −4 ) + ( −2 )
∫∫ yz + 4 xy dS + 1 dA
2 2
S1 D
= 21 ∫∫ 8 y − 2 y 2 dA
D
Don’t forget to plug the equation of the surface into z in the integrand and don’t forget to use the equation
of the surface in the computation of the root!
Now, as noted above D for this surface is nothing more than S 4 and so we can use the limits from the
sketch of S 4 in Step 1.
Now let’s compute the integral for this surface.
2 4− 2 x
∫∫ yz + 4=
S1
xy dS 21 ∫
0 ∫ 0
8 y − 2 y 2 dy dx
2 4− 2 x
= 21⌠  4 y 2 − 32 y 3  dx
⌡0 0
21 ∫ 4 ( 4 − 2 x ) − 32 ( 4 − 2 x ) dx
2
=
2 3
0
2
= 21  − 23 ( 4 − 2 x ) + 121 ( 4 − 2 x )  = = 97.7616
3 4 64 21
 0 3
Step 3
Next we’ll take care of S 2 . In this case the equation for the surface is simply y = 0 and D is
given in the sketch of S 2 in Step 1.
dS ∫∫ 0 ( 0 )
∫∫ yz + 4 xy= + 1 + ( 0 )= ∫∫ 0=
2 2
dA dA 0
S2 D D
So, in this case we didn’t need to actually compute the integral. Sometimes we’ll get lucky like this,
although it probably won’t happen all that often.
Step 4
Now we can take care of S3 . In this case the equation for the surface is simply x = 0 and D is given in
the sketch of S3 in Step 1.
∫∫ yz + 4 xy dS = ∫∫  yz + 4 ( 0 ) y  1 + ( 0 ) + ( 0 ) dA= ∫∫ yz dA
2 2
S3 D D

Calculus II
Don’t forget to plug the equation of the surface into x in the integrand and don’t forget to use the equation
of the surface in the computation of the root (although in this case the root just evaluates to one)!
Using the limits for D from the sketch in Step 1 we can quickly evaluate the integral for this surface.
4 8− 2 y
∫∫ yz + 4 xy dS =
S3
∫ ∫ yz dz dy 0 0
4 8− 2 y
= ⌠  12 yz 2  dy
⌡0 0
4
= ∫0
32 y − 16 y 2 + 2 y 3 dy
4
=16 y 2 − 163 y 3 + 12 y 4  =128
3 =42.6667
0
Step 5
Finally let’s take care of S 4 . In this case the equation for the surface is simply z = 0 and D is given in
the sketch of S 4 in Step 1.
= ∫∫  y ( 0 ) + 4 xy  ( 0 ) + ( 0 )
∫∫ yz + 4 xy dS + 1 dA
= ∫∫ 4 xy dA
2 2
S4 D D
Don’t forget to plug the equation of the surface into z in the integrand and don’t forget to use the equation
of the surface in the computation of the root (although in this case the root just evaluates to one)!
Using the limits for D from the sketch in Step 1 we can quickly evaluate the integral for this surface.
2 4− 2 x
∫∫ yz + 4 xy dS =
S4
∫ ∫ 4 xy dy dx 0 0
2 4− 2 x
= ⌠  2 xy 2  dx
⌡0 0
2
= ∫ 0
32 x − 32 x 2 + 8 x3 dx
2
=16 x − 3 x + 2 x  0 =3 =
2 32 3 4 32
10.6667
Step 6
Now, to get the value of the integral over the full surface all we need to do is sum up the values of each of
the integrals over the four surfaces above. Doing this gives,
∫∫ yz + 4 xy dS = (
S
64 21
3 ) + ( 0) + ( 128
3 ) + ( 323 )= 64 21
3 + 160
3 = 151.0949
We put parenthesis around each of the individual integral values just to indicate where each came from.
In general these aren’t needed of course.
Calculus II
∫∫ x − z dS where S is the surface of the solid bounded by x 4 , z= x − 3 , and

+ y2 =
2
6. Evaluate
S
z= x + 2 . Note that all three surfaces of this solid are included in S.
Step 1

Calculus II
As noted in the problem statement there are three surfaces here. The “top” of the cylinder is a little hard
to see. We made the walls of the cylinder slightly transparent and the top of the cylinder can be seen as a
darker ellipse along the top of the surface.
To help visualize the relationship between the top and bottom of the cylinder here is a different view of
the surface.

Calculus II
From this view we can see that the top and bottom planes that “cap” the cylinder are parallel.
Let’s define the three surfaces in the sketch as follows.
4 ( i.e the walls of the solid )

S1 : Cylinder given by x 2 + y 2 =
S 2 : Plane given by z= x + 2 ( i.e the top "cap" of the cylinder )
S3 : Plane given by z= x − 3 ( i.e the bottom "cap" of the cylinder )
Step 2
Let’s start with S1 . The surface in this case is a cylinder and so we’ll need to parameterize it.
The parameterization of the surface is,

r ( z , θ ) = 2 cos θ , 2sin θ , z
The limits on z and θ are,

Calculus II
0 ≤ θ ≤ 2π , 2 cos θ − 3 = x − 3 ≤ z ≤ x + 2 = 2 cos θ + 2
With the z limits we’ll need to make sure that we convert the x’s into their parameterized form.
 
In order to evaluate the integral in this case we’ll need the cross product rz × rθ so here is that work.
 
rz = 0, 0,1 rθ = −2sin θ , 2 cos θ , 0
  
i j k
   
rz × rθ = 0 0 1 =−2 cos θ i − 2sin θ j
−2sin θ 2 cos θ 0
Next we’ll need the magnitude of the cross product so here is that.
 
rz=
× rθ 4 cos 2 θ + 4sin
= 2
θ 2
dS ∫∫ [ 2 cos θ − z ] ( 2 ) =
∫∫ x − z = dA ∫∫ 4 cos θ − 2 z dA
S1 D D
Don’t forget to plug the parameterization of the surface into the integrand and don’t forget to add in the
magnitude of the cross product!
Now, D for this surface is nothing more than the limits on z and θ we gave above.
Now let’s compute the integral for this surface.
2π 2cosθ + 2
∫∫ x − z dS
=
S1
∫ ∫
0 2cosθ −3
4 cos θ − 2 z dz dθ
2π
( 4 z cos θ − z )
2cosθ + 2
= ⌠ 2
dθ
⌡0 2cosθ −3
2π
= ∫ 4 cos θ  2 cos θ + 2 − ( 2 cos θ − 3)  − ( 2 cos θ + 2 ) − ( 2 cos θ − 3)  dθ
2 2
0  
2π
= ∫=
5 dθ
0
10π
Do not forget to simplify! As we saw with this problem after the z integration the integrand looked really
messy but after some pretty simple simplification it reduced down to an incredibly simple integrand.
Step 3
Next we’ll take care of S 2 . In this case the equation for the surface is simply z= x + 2 and D is
the disk x 2 + y 2 ≤ 4 .

Calculus II
∫∫ x − z dS =∫∫  x − ( x + 2 ) (1) + ( 0 ) + 1 dA =∫∫ −2 2 dA =−2 2 ∫∫ dA

2 2
S2 D D D
Okay, in this case we don’t need to actually do the evaluation of the integral because we know that,
∫∫ dA = Area of D
D
and in this case D is just a disk and we can quickly determine its area without any evaluation.
So, the integral for this surface is then just,
∫∫ x − z dS = −2 2 ( 2 )
−2 2 ( Area of D ) = π =
−8 2π
2
 
S2
Step 4
Finally, let’s integrate over S3 . In this case the equation for the surface is simply z= x − 3 and D is the
disk x 2 + y 2 ≤ 4 .
∫∫ x − z dS= ∫∫  x − ( x − 3) (1) + ( 0 ) + 1 dA

2 2
S3 D
= ∫∫=
3 2 dA 3= 2 ( 4π ) 12 2π
2 ∫∫ dA 3=
D D
So, the integral in this case ended up being every similar to the integral in Step 3 and so we didn’t put in
any of the explanation here.
Step 5
the integrals over the three surfaces above. Doing this gives,
∫∫ x − z dS= (10π ) + ( −8
S
) (
2π + 12 2π = ) (10 + 4 2 ) π= 49.1875

Calculus II
Surface Integrals of Vector Fields
     
1. Evaluate ∫∫ F  dS where F= 3 x i + 2 z j + (1 − y 2
) k and S is the portion of z =2 − 3 y + x 2 that lies
S
over the triangle in the xy-plane with vertices ( 0, 0 ) , ( 2, 0 ) and ( 2, −4 ) oriented in the negative z-axis
direction.
Step 1
surface.

Calculus II
The orange surface is the sketch of z =2 − 3 y + x 2 that we are working with in this problem. The
greenish triangle below the surface is the triangle referenced in the problem statement that lies below the
surface. This triangle will be the region D for this problem.
Here is a quick sketch of D just to get a better view of it than the mostly obscured view in the sketch
above.
We could use either of the following sets of limits to describe D.
0≤ x≤2 −4≤ y ≤0
−2 x ≤ y ≤ 0 − 12 y ≤ x ≤ 2
We’ll decide which set to use in the integral once we get that set up.
Step 2
Let’s get the integral set up now. In this case the we can write the equation of the surface as
follows,
f ( x, y , z ) = 2 − 3 y + x 2 − z = 0
A unit normal vector for the surface is then,
 ∇f 2 x, −3, −1
=n =
∇f ∇f
We didn’t compute the magnitude of the gradient since we know that it will just cancel out when we start
working with the integral.
Note as well that, in this case, the normal vector we computed above has the correct orientation. We were
told in the problem statement that the orientation was in the negative z-axis direction and this means that
the normal vector should always have a downwards direction (i.e. a negative z component) and this one
does.
Step 3
Next, we’ll need to compute the following dot product.

Calculus II
 2 x, −3, −1
)n
F ( x, y, 2 − 3 y + x 2= 3 x, 2 ( 2 − 3 y + x 2 ) ,1 − y 2 
∇f
= 1
∇f (6x 2
− 6 ( 2 − 3 y + x 2 ) − (1 − y 2 ) )
= 1
∇f (y 2
+ 18 y − 13)
Remember that we needed to plug in the equation of the surface, z =2 − 3 y + x 2 , into z in the vector
field!

 
∫∫ = ∫∫ ( y + 18 y − 13) dS
1 2
F dS ∇f
S S
= ∫∫ ( y + 18 y − 13) ∇f dA
1 2
∇f
D
= ∫∫ y + 18 y − 13 dA
2
As noted above we didn’t need to compute the magnitude of the gradient since it would just cancel out
when we converted the surface integral into a “normal” double integral.
Also, recall that D was given in Step 1. We had two sets of limits to use here but it seems like the first set
is probably just as easy to use so we’ll use that one in the integral.
Step 4
Here is the integral,

 
∫∫  dS = ∫∫ y + 18 y − 13 dA
2
F
S D
2 0
= ∫∫
0 −2 x
y 2 + 18 y − 13 dy dx
= ⌠ ( 13 y 3 + 9 y 2 − 13 y ) dx
2 0
⌡0 −2 x
2
= ∫ 8
0 3
x 3 − 36 x 2 − 26 x dx
( 23 x 4 − 12 x3 − 13x 2 ) =
2
= − 412
3
0

Calculus II
     
2. Evaluate ∫∫
S
F  dS where F =
− x i + 2 y j − z k and S is the portion of=
y 3 x 2 + 3 z 2 that lies behind
y = 6 oriented in the positive y-axis direction.
Step 1

Calculus II
Note that the surface in this problem is only the elliptic paraboloid and does not include the “cap” at
y = 6 . We would only include the “cap” if the problem had specified that in some manner to make it
clear.
In this case D will be the circle/disk we get by setting the two equations equal or,
6 = 3x 2 + 3z 2 ⇒ x2 + z 2 = 2
So, D will be the disk x 2 + z 2 ≤ 2 .
Step 2
Let’s get the integral set up now. In this case the we can write the equation of the surface as
follows,
f ( x, y, z )= 3 x 2 + 3 z 2 − y= 0
 ∇f 6 x, −1, 6 z
=n =
∇f ∇f
Note as well that, in this case, the normal vector we computed above does not have the correct
orientation. We were told in the problem statement that the orientation was in the positive y-axis
direction and this means that the normal vector should always point in the general direction of the positive
y-axis (i.e. a positive y component) and this one does not.
That is easy to fix however. All we need to do is multiply the above normal vector by minus one and
we’ll get what we need. So, here is the normal vector we need for this problem.
 ∇f −6 x,1, −6 z
n=− =
∇f ∇f
As we can see this normal vector does in fact have a positive y component as we need.
Step 3
  −6 x,1, −6 z
F ( x,3 x 2 + 3 z 2 , z )n =− x, 2 ( 3 x 2 + 3 z 2 ) , − z 
∇f
= 1
∇f (6x 2
+ 2 ( 3x 2 + 3z 2 ) + 6 z 2 )
= 1 12 ( x 2 + z 2 ) 
∇f  

Calculus II
Remember that we needed to plug in the equation of the surface,=

y 3 x 2 + 3 z 2 , into y in the vector field!

 
∫∫  dS
= F ∫∫ 1 12 ( x 2 + z 2 )  dS
∇f  
S S
= ∫∫ 1 12 ( x 2 + z 2 )  ∇f dA
∇f  
D
= ∫∫ 12 ( x + z 2 ) dA
2
As noted above we didn’t need to compute the magnitude of the gradient since it would just cancel out
when we converted the surface integral into a “normal” double integral.
Also, recall that D was given in Step 1 and is just the disk x + z ≤ 2
2 2
Step 4
Note as well that we’ll want to use polar coordinates in the double integral. We’ll use the following set of
polar coordinates.
= θ
The polar limits for D are,
0 ≤ θ ≤ 2π
0≤r ≤ 2

 
F  dS ∫∫ 12 ( x
∫∫=
2
+ z 2 ) dA
S D
2π
=⌠
2
⌡0 ∫0
12r 3 dr dθ
2π
= ⌠ 3r 4 dθ
2
⌡0 0
2π
= ∫=
0
12 dθ 24π
Don’t forget that we pick up an extra r from the dA when converting to polar coordinates.

Calculus II
     
∫∫  = + − 4 between x = 0
and S is the portion of y 2 + z 2 =
2
3. Evaluate F dS where F x i 2 z j 3 y k
S
and x= 3 − z oriented outwards (i.e. away from the x-axis).
Step 1
Note that the surface in this problem is only the cylinder itself. The “caps” of the cylinder are not part of
this surface despite the red “cap” in the sketch. That was included in the sketch to make the front edge of
the cylinder clear in the sketch. We would only include the “caps” if the problem had specified that in
some manner to make it clear.

Calculus II
Step 2
Let’s get the integral set up now. In this case the we are integrating over a cylinder and so we’ll
need to set up a parameterization for the surface.
We saw how to parameterize a cylinder in the first section of this chapter so we won’t go into detail for
the parameterization. The parameterization is,

r ( x, θ ) = x, 2sin θ , 2 cos θ 0 ≤ θ ≤ 2π , 0 ≤ x ≤ 3 − z = 3 − 2 cos θ
We’ll use the full range of θ since we are allowing it to rotate all the way around the x-axis. The x limits
come from the two planes that “bound” the cylinder and we’ll need to convert the upper limit using the
parameterization.

 
=rx 1, 0, 0 =rθ 0, 2 cos θ , −2sin θ
  
i j k
   
rx =
× rθ 1 0 0= 2sin θ j + 2 cos θ k
0 2 cos θ −2sin θ

 
 rx × rθ 0, 2sin θ , 2 cos θ
=n =
   
rx × rθ rx × rθ
We didn’t compute the magnitude of the cross product since we know that it will just cancel out when we
start working with the integral.
Now we need to determine if this vector has the correct orientation. First let’s look at the cylinder from in
front of the cylinder and directly along the x-axis. This is what we’d see.

Calculus II
In this sketch the x axis will be coming straight out of the sketch at the origin. Plugging in a few value of
θ into the parameterization we can see that we’ll be at the points listed above.
Now, in the range 0 ≤ θ ≤ 12 π we know that sine and cosine are both positive and so in the normal vector
both the y and z components will be positive. This means that in the 1st quadrant above the normal vector
would need to be pointing out away from the origin. This is exactly what we need to see since the
orientation was given as pointing away from the x-axis and recall that the x-axis is coming straight out of
the sketch from the origin.
Next, if we look at 12 π ≤ θ ≤ π (so we’re in the 4th quadrant of the graph above….) we know that in this
range sine is still positive but cosine is now negative. From our unit vector above this means that the y
component is positive (so pointing in positive y direction) and the z component is negative (so pointing in
negative z direction). Together this again means that we have to be pointing away from the origin in the
4th quadrant which is again the orientation we want.
We could continue in this fashion looking at the remaining two quadrants but once we’ve done a couple
and gotten the correct orientation we know we’ll continue to get the correct orientation for the rest.
Step 3
   0, 2sin θ , 2 cos θ
F ( r ( x, θ ) )n
= x 2 , 2 ( 2 cos θ ) , −3 ( 2sin θ )   
rx × rθ
=  
1
rx ×rθ ( −4sin θ cos θ )
Remember that we needed to plug in the parameterization for the surface into the vector field!

Calculus II
 
∫∫=
F  dS ∫∫  
1
rx ×rθ ( −4sin θ cos θ ) dS
S S
 
∫∫ rx ×1rθ ( −4sin θ cos θ ) rx × rθ dA
=
D
= ∫∫ −4sin θ cos θ dA
D
As noted above we didn’t need to compute the magnitude of the cross product since it would just cancel
out when we converted the surface integral into a “normal” double integral.
Also, recall that D is given by the limits on x and θ we found at the start of Step 2.
Step 4

 
∫∫ F  dS= ∫∫ −4sin θ cos θ dA
S D
2π 3− 2cosθ
= ∫ ∫
0 0
−4sin θ cos θ dx dθ
2π 3− 2cosθ
= ∫ −4 x sin θ cos θ
0
dθ0
π
= ∫ −4 ( 3 − 2 cos θ ) sin θ cos θ dθ
2
0
π 2
= ∫ −12sin θ cos θ + 8sin θ cos θ dθ
2
0
π
∫ −6sin ( 2θ ) + 8sin θ cos θ dθ
2
= 2
0
( −3cos ( 2θ ) − 83 cos3 θ )
2π
= =
0
0
     
4. Evaluate ∫∫
S
F  dS where F =+
i z j + 6 x k and S is the portion of the sphere of radius 3 with x ≤ 0 ,
y ≥ 0 and z ≥ 0 oriented inward (i.e. towards the origin).
Step 1

Calculus II
Note that the surface in this problem is only the part of the sphere itself. The “edges” (the greenish
portions on the right/left) are not part of this surface despite the fact that they are in the sketch. They

Calculus II
were included in the sketch to try and make the surface a little clearer in the sketch. We would only
include the “edges” if the problem had specified that in some manner to make it clear.
Step 2
Let’s get the integral set up now. In this case the we are integrating over a sphere and so we’ll
need to set up a parameterization for the surface.
We saw how to parameterize a sphere in the first section of this chapter so we won’t go into detail for the

r (θ , ϕ ) 3sin ϕ cos θ ,3sin ϕ sin θ ,3cos ϕ 1
2 π ≤ θ ≤ π , 0 ≤ ϕ ≤ 12 π
We needed the restriction on ϕ to make sure that we only get a portion of the upper half of the sphere
(i.e. z ≥ 0 ). Likewise the restriction on θ was needed to get only the portion that was in the 2nd quadrant
of the xy-plane (i.e. x ≤ 0 and y ≥ 0 ).

 
rϕ =
  
i j k
 
   
−9 sin 2 ϕ cos θ i − 9 sin ϕ cos ϕ sin 2 θ k − 9 sin ϕ cos ϕ cos 2 θ k − 9 sin 2 ϕ sin θ j
=
  
=
  
=
 
 rθ × rϕ −9sin 2 ϕ cos θ , −9sin 2 ϕ sin θ , −9sin ϕ cos ϕ
=n =
   
rθ × rϕ rθ × rϕ
Now we need to determine if this vector has the correct orientation. We know that the normal vector
needs to point in towards the origin. Let’s think about what that would mean for a normal vector on the
upper half of a sphere and it won’t matter which quadrant in the xy-plane we are in.
If we are on the upper half of a sphere and the normal vectors must point towards the origin then we know
that they will all need to point downwards. They could point in the positive or negative x (or y) direction
depending on which quadrant from the xy-plane we are on but they will have to all point downwards. Or
in other words, the z component must be negative.

Calculus II
So, the z component of the normal vector above is −9sin ϕ cos ϕ and we know that we are restricted to
0 ≤ ϕ ≤ 12 π for the portion of the sphere we are working on in this problem. In this range of ϕ we know
that both sine and cosine are positive and so the z component must always be negative. This means that
the normal vector above has the correct orientation for this problem.
Note that if we were on the lower half of a sphere (not relevant for this problem but useful to think about
anyway) and the normal vector would be pointing towards the origin and so they would have to all be
pointing upwards.
Also note that if the normal vectors were all pointing out away from the origin then we’d just need to
multiply the normal vector above by minus one to get the normal vector we’d need.
Step 3
   −9sin 2 ϕ cos θ , −9sin 2 ϕ sin θ , −9sin ϕ cos ϕ

F ( r (θ , ϕ ) )n = 1,3cos ϕ ,18sin ϕ cos θ   
rθ × rϕ
r ×r (
=−1
  9sin 2 ϕ cos θ − 27 sin 2 ϕ cos ϕ sin θ − 162sin 2 ϕ cos ϕ cos θ )
θ ϕ
=  
1
rθ ×rϕ ( − (1 − cos ( 2ϕ ) ) cosθ − sin
9
2
2
ϕ cos ϕ ( 27 sin θ + 162 cos θ ) )
Note that we did a little simplification for the integration process in the last step above.

 
∫∫  =
F dS ∫∫  
1
9
2
2
ϕ cos ϕ ( 27 sin θ + 162 cos θ ) ) dS
S S
 
= ∫∫  
1
9
2
2
ϕ cos ϕ ( 27 sin θ + 162 cos θ ) ) rθ × rϕ dA
D
=∫∫ − 92 (1 − cos ( 2ϕ ) ) cos θ − sin 2 ϕ cos ϕ ( 27 sin θ + 162 cos θ ) dA

D
As noted above we didn’t need to compute the magnitude of the cross product since it would just cancel
out when we converted the surface integral into a “normal” double integral.
Also, recall that D is given by the limits on θ and ϕ we found at the start of Step 2.
Step 4

Calculus II
 
∫∫ F  dS =∫∫ − (1 − cos ( 2ϕ ) ) cos θ − sin ϕ cos ϕ ( 27 sin θ + 162 cos θ ) dA
9 2
2
S D
π
= ⌠ ∫ − 92 (1 − cos ( 2ϕ ) ) cos θ − sin 2 ϕ cos ϕ ( 27 sin θ + 162 cos θ ) dϕ dθ
1π
2
⌡2π 0
1
π
= ⌠ − 92 (ϕ − 12 sin ( 2ϕ ) ) cos θ − 13 sin 3 ϕ ( 27 sin θ + 162 cos θ ) dθ
1π
2
⌡ 12 π 0
π
∫1 − 94 π cos θ − 9sin θ − 54 cos θ dθ
=
2
π
π
( − 94 π sin θ + 9 cos θ − 54sin θ ) 1 π =
= 4 π + 45
9
2
     
5. Evaluate ∫∫ F  dS where F = y i + 2 x j + ( z − 8 ) k and S is the surface of the solid bounded by
S
4x + 2 y + z =8 , z = 0 , y = 0 and x = 0 with the positive orientation. Note that all four surfaces of
this solid are included in S.
Step 1

Calculus II
Okay, as noted in the problem statement all four surfaces in the sketch (two not shown) are part of S so
let’s define each of them as follows.
S1 : Plane given by 4 x + 2 y + z =8 ( i.e the top of the solid )

S 2 : Plane given by y = 0 ( i.e the triangle on right side of the solid )
S3 : Plane given by x = 0 ( i.e the triangle at back of the solid - not shown in sketch )
S 4 : Plane given by z = 0 ( i.e the triangle on bottom of the solid - not shown in sketch )
As noted in the definitions above the first two surfaces are shown in the sketch but the last two are not
actually shown due to the orientation of the solid. Below are sketches of each of the three surfaces that
correspond to the coordinates planes.

Calculus II
With each of the sketches we gave limits on the variables for each of them since we’ll eventually need
that when we start doing the surface integral along each surface.
Step 2
Let’s start with S1 . In this case we can write the equation of the plane as follows,
f ( x, y , z ) = 4 x + 2 y + z − 8 = 0
 ∇f 4, 2,1
=n =
∇f ∇f

Calculus II
The surface has the positive orientation and so must point outwards from the region enclosed by
the surface. This means that normal vectors on this plane will need to be pointing generally
upwards (i.e. have a positive z component) which this normal vector does.
Now we’ll need the following dot product and don’t forget to plug in the equation of the plane
(solved for z of course) into z in the vector field.
  4, 2,1
F ( x, y,8 − 4 x − 2 y )=
n y, 2 x, −4 x − 2 y 
∇f
= 1
∇f (2y)
 
∫∫  dS = ∫∫
F 1
∇f ( 2 y ) dS
S1 S1
= ∫∫ ( 2 y ) ∇f
D
1
∇f
dA
= ∫∫ 2 y dA
D
In this case D is just S 4 and so we can now finish out the integral.
 
∫∫ F  dS = ∫∫ 2 y dA
S1 D
2 4− 2 x
=∫ ∫ 2 y dy dx
0 0
2 4− 2 x
= ∫ y2 dx
0 0
∫ ( 4 − 2x)
2
=
2
dx
0
3 2
− 16 ( 4 − 2 x )
= =
32
3
0
Step 3
Next we’ll take care of S 2 . In this case the equation for the surface is simply y = 0 and D is given in the
sketch of S 2 in Step 1.
In this case S 2 is simply a portion of the xz-plane and we have the positive orientation and so the normal
vector must point away from the region enclosed by the surface. That means that, in this case, the normal
 
vector is simply n =− j =0, −1, 0 .
The dot product for this surface is,

Calculus II
  0, −1, 0
F ( x, 0, z )n =
0, 2 x, z − 8  =
−2 x
1
Don’t forget to plug y = 0 into the vector field and note that the magnitude of the gradient is,
∇= ( 0 ) + (1) + ( 0=
)
2 2 2
f 1

 
∫∫  dS=
S2
F ∫∫ −2 x dS
S2
= ∫∫ −2 x (1) dA
D
2 8− 4 x
= ∫ ∫ −2 x dz dx
0 0
2 8− 4 x
= ∫ −2 xz dx
0 0
2
= ∫ 8 x − 16 x dx
2
0
( 83 x3 − 8 x 2 ) =
2
= − 323
0
Step 4
Now we can deal with S3 . In this case the equation for the surface is simply x = 0 and D is given in the
sketch of S3 in Step 1.
In this case S3 is simply a portion of the yz-plane and we have the positive orientation and so the normal
 
vector is simply n =−i = −1, 0, 0 .
  −1, 0, 0
F ( 0, y, z )n =
y, 0, z − 8  =
−y
1
Don’t forget to plug x = 0 into the vector field and note that the magnitude of the gradient is,
∇f= (1) + ( 0 ) + ( 0 )=
2 2 2
1

Calculus II
 
∫∫ F  dS= ∫∫ − y dS
S3 S3
= ∫∫ − y (1) dA
D
4 8− 2 y
= ∫ ∫ − y dz dy
0 0
4 8− 2 y
= ∫ − yz dy
0 0
4
= ∫ 2 y − 8 y dy
2
0
( 23 y3 − 4 y 2 ) =
4
= − 643
0
Step 5
Finally let’s take care of S 4 . In this case the equation for the surface is simply z = 0 and D is given in
the sketch of S 4 in Step 1.
In this case S 4 is simply a portion of the xy-plane and we have the positive orientation and so the normal
 
vector is simply n =− k =0, 0, −1 .
  0, 0, −1
F ( x , y , 0 ) n =
y, 2 x, −8  =
8
1
Don’t forget to plug z = 0 into the vector field and note that the magnitude of the gradient is,
∇= ( 0 ) + ( 0 ) + (1=
)
2 2 2
f 1

 
∫∫ F  dS = ∫∫ 8 dS
S4 S4
= ∫∫ 8 dA
D
= 8∫∫ dA
D
= 8 ( Area of D )
( 12 ) ( 2 )( 4 ) 32
= 8=
In this case notice that we didn’t have to actually compute the double integral since D was just a right
triangle and we can easily compute its area.

Calculus II
Step 6
the integrals over the four surfaces above. Doing this gives,
 
∫∫  dS=
F ( 323 ) + ( − 323 ) + ( − 643 ) + ( 32=) 32
3
S
     
6. Evaluate ∫∫ F  dS
S
where F = yz i + x j + 3 y 2 k and S is the surface of the solid bounded by
2
4 , z= x − 3 , and z= x + 2 with the negative orientation. Note that all three surfaces of this
x +y = 2
solid are included in S.
Step 1

Calculus II
As noted in the problem statement there are three surfaces here. The “top” of the cylinder is a little hard
to see. We made the walls of the cylinder slightly transparent and the top of the cylinder can be seen as a
darker ellipse along the top of the surface.
To help visualize the relationship between the top and bottom of the cylinder here is a different view of
the surface.

Calculus II
From this view we can see that the top and bottom planes that “cap” the cylinder are parallel.
Let’s define the three surfaces in the sketch as follows.
4 ( i.e the walls of the solid )

S1 : Cylinder given by x 2 + y 2 =
S 2 : Plane given by z= x + 2 ( i.e the top "cap" of the cylinder )
S3 : Plane given by z= x − 3 ( i.e the bottom "cap" of the cylinder )
Step 2
Let’s start with S1 . The surface in this case is a cylinder and so we’ll need to parameterize it.
The parameterization of the surface is,

r ( z , θ ) = 2 cos θ , 2sin θ , z
The limits on z and θ are,

Calculus II
0 ≤ θ ≤ 2π , 2 cos θ − 3 = x − 3 ≤ z ≤ x + 2 = 2 cos θ + 2
With the z limits we’ll need to make sure that we convert the x’s into their parameterized form.
 
In order to evaluate the integral in this case we’ll need the cross product rz × rθ so here is that work.
 
rz = 0, 0,1 rθ = −2sin θ , 2 cos θ , 0
  
i j k
   
rz × rθ = 0 0 1 =−2 cos θ i − 2sin θ j
−2sin θ 2 cos θ 0

 
 rz × rθ −2 cos θ , −2sin θ , 0
=n =
   
rz × rθ rz × rθ
The surface has the negative orientation and so must point in towards the region enclosed by the
surface. This means that normal vectors on cylinder will need to point in towards the z-axis and
this vector does point in that direction.
To see that this vector points in towards the z-axis consider the 0 ≤ θ ≤ π2 . In this range both sine and
cosine are positive and so the x and y component of the normal vector will be negative and so will point in
towards the z-axis.
Now we’ll need the following dot product and don’t forget to plug in the parameterization of the
surface in the vector field.
   −2 cos θ , −2sin θ , 0
F ( r ( z , θ ) )n = 2 z sin θ , 2 cos θ ,12sin 2 θ   
rz × rθ
rz ×rθ (
=
 
1
−4 z sin θ cos θ − 4sin θ cos θ )
   −4sin θ cos θ ( z + 1) 
= 1
rz ×rθ  

Calculus II
 
F
S1
∫∫ rz ×1rθ −2sin ( 2θ )( z + 1) dS
∫∫  dS = S1
 
= ∫∫  
1
rz ×rθ
 −2sin ( 2θ )( z + 1)  rz × rθ dA
D
∫∫ −2sin ( 2θ )( z + 1) dA
=
D
In this case D is is nothing more than the limits on z and θ we gave above and so we can now finish out
the integral.
 
S1
∫∫ −4sin θ cos θ ( z + 1) dA
∫∫ F  dS = D
2π 2cosθ + 2
∫ ∫
= −4sin θ cos θ ( z + 1) dz dθ
0 2cosθ −3
2π
⌠ −4sin θ cos θ ( 1 z 2 + z )
2cosθ + 2
= dθ
⌡0 2
2cosθ −3
2π
∫ −10sin θ cos θ − 40sin θ cos θ dθ
= 2
0
2π
∫ −5sin ( 2θ ) − 40sin θ cos θ dθ
= 2
0
3 2π
=( 52 cos ( 2θ ) + 403 cos3 θ ) 0
=0
Step 3
Next we’ll take care of S 2 . In this case the equation for the surface is can be written as z − x − 2 =0
and D is the disk x 2 + y 2 ≤ 4 .
A unit normal vector for S 2 is then,
 ∇f −1, 0,1
=n =
∇f ∇f
The region has the negative orientation and so must point into the enclosed region and so must point
downwards (since this is the top “cap” of the cylinder). The normal vector above points upwards (it has a
positive z component) and so we’ll need to multiply this by minus one to get the normal vector we need
for this surface.
The correct normal vector is then,
 ∇f 1, 0, −1
n=− =
∇f ∇f

Calculus II
The dot product we’ll need for this surface is,
  1, 0, −1
F ( x, y, x + 2 )n = y ( x + 2 ) , x,3 y 2 
∇f
= 1
∇f ( xy + 2 y − 3 y ) 2
Don’t forget to plug the equation of the surface into z in the vector field.

 
∫∫ = dS ∫∫ ( xy + 2 y − 3 y ) dS
1 2
F ∇f
S2 S2
= ∫∫ ( xy + 2 y − 3 y ) ∇f
1 2
∇f
dA
D
= ∫∫ xy + 2 y − 3 y
2
dA
D
Note that we’ll need to finish this integral with polar coordinates and the polar limits will be,
0 ≤ θ ≤ 2π
0≤r ≤2

 
∫∫ F  dS= ∫∫ xy + 2 y − 3 y
2
dA
S2 D
∫ ∫ ( r sin θ cos θ + 2r sin θ − 3r sin θ ) ( r ) dr dθ

2π 2
= 2 2 2
0 0
= ∫ ∫ r sin ( 2θ ) + 2r sin θ − r (1 − cos ( 2θ ) ) dr dθ

π 2 2
1 3 2 3 3
0 0 2 2
r sin ( 2θ ) + r sin θ − r (1 − cos ( 2θ ) ) dθ

π 2
= ∫
2
1 4 2 3 3 4
0 8 3 8 0
= ∫ 2sin ( 2θ ) + sin θ − 6 (1 − cos ( 2θ ) ) dθ

π 2
16
0 3
( − cos ( 2θ ) − cosθ − 6 (θ − sin ( 2θ ) ) ) = −12π

π 2
= 16
3
1
2
0
Step 4
Finally, let’s integrate over S3 . In this case the equation for the surface is can be written as z − x + 3 =0
and D is the disk x 2 + y 2 ≤ 4 .
A unit normal vector for S 2 is then,

Calculus II
 ∇f −1, 0,1
=n =
∇f ∇f
The region has the negative orientation and so must point into the enclosed region and so must point
upwards (since this is the bottom “cap” of the cylinder). The normal vector above does point upwards (it
has a positive z component) and so is the normal vector we’ll need.
The dot product we’ll need for this surface is,
  −1, 0,1
F ( x , y , x − 3 ) n = y ( x − 3) , x,3 y 2 
∇f
= 1
∇f ( − xy + 3 y + 3 y )2
Don’t forget to plug the equation of the surface into z in the vector field.

 
= ∫∫ ( − xy + 3 y + 3 y ) dS
∫∫ F  dS 1
∇f
2
S3 S3
= ∫∫ ( − xy + 3 y + 3 y ) ∇f
1 2
∇f
dA
D
= ∫∫ − xy + 3 y + 3 y 2 dA
D
Note that we’ll need to finish this integral with polar coordinates and the polar limits will be,
0 ≤ θ ≤ 2π
0≤r ≤2

Calculus II
 
∫∫  dS = ∫∫ − xy + 3 y + 3 y dA
2
F
S3 D
∫ ( −r sin θ cos θ + 3r sin θ + 3r sin θ ) ( r ) dr dθ

2π 2
=∫ 2 2 2
0 0
= ∫ ∫ − r sin ( 2θ ) + 3r sin θ + r (1 − cos ( 2θ ) ) dr dθ

π 2 2
1 3 2 3 3
0 0 2 2
∫ − r sin ( 2θ ) + r sin θ + r (1 − cos ( 2θ ) ) dθ

π 2 2
= 1
8
4 3 3
8
4
0 0
= ∫ −2sin ( 2θ ) + 16sin θ + 6 (1 − cos ( 2θ ) ) dθ

π 2
= ( cos ( 2θ ) − 16 cos θ + 6 (θ − sin ( 2θ ) ) ) = 12π

π 2
1
2
0
Step 5
the integrals over the three surfaces above. Doing this gives,
 
∫∫  dS=
F ( 0 ) + ( −12π ) + (12π )= 0
S
Stokes’ Theorem
     
∫∫  = − + 3
1. Use Stokes’ Theorem to evaluate curl F dS where F y i x j yx k and S is the portion of
S
the sphere of radius 4 with z ≥ 0 and the upwards orientation.
Step 1

Calculus II
surface.
Because the orientation of the surface is upwards then all the normal vectors will be pointing outwards.
So, if we walk along the edge of the surface, i.e. the curve C, in the direction indicated with our head
pointed away from the surface (i.e. in the same direction as the normal vectors) then our left hand will be
over the region. Therefore the direction indicated in the sketch is the positive orientation of C.
If you have trouble visualizing the direction of the curve simply get a cup or bowl and put it upside down
on a piece of paper on a table. Sketch a set of axis on the piece of paper that will represent the plane the
cup/bowl is sitting on to really help with the visualization. Then cut out a little stick figure and put a face
on the “front” side of it and color the left hand a bright color so you can quickly see it. Now, on the edge
of the cup/bowl/whatever you place the stick figure with it’s head pointing in the direction of the normal
vectors (out away from the sphere/cup/bowl in our case) with its left hand over the surface. The direction
that the “face” on the stick figure is facing is the direction you’d need to walk along the surface to get the
positive orientation for C.
Step 2

Calculus II
We are going to use Stokes’ Theorem in the following direction.

   
∫∫
S
curl F  dS = ∫ dr
F
C
We’ve been given the vector field in the problem statement so we don’t need to worry about that. We
will need to deal with C.
Because C is just the curve along the bottom of the upper half of the sphere we can see that C in fact will
be the intersection of the sphere and the xy-plane (i.e. z = 0 ). Therefore, C is just the circle of radius 4.
If we look at the sphere from above we get the following sketch of C.
The parameterization of C is given by,


=r (t ) 4 cos t , 4sin t , 0 0 ≤ t ≤ 2π
The z component of the parameterization is zero because C lies in the xy-plane.
Step 3
Since we know we’ll need to eventually do the line integral we know we’ll need the following dot
product.
  
F ( r ( t ) )r ' ( t ) =
4sin t , −4 cos t , 256sin t cos3 t  −4sin t , 4 cos t , 0
= −16sin 2 t − 16 cos 2 t
= −16
Don’t forget to plug the parameterization of C into the vector field!
Step 4
Okay, let’s go ahead and evaluate the integral using Stokes’ Theorem.

Calculus II
   
∫∫
S
curl F  dS = ∫ dr
F
C
2π
∫ 16 dt =
=−
0
−32π
     
2. Use Stokes’ Theorem to evaluate ∫∫ curl F dS where F = (z 2
− 1) i + ( z + xy 3 ) j + 6 k and S is the
S
portion of x =−
6 4 y − 4z
2 2
in front of x = −2 with orientation in the negative x-axis direction.
Step 1

Calculus II
surface.
Because the orientation of the surface is towards the negative x-axis all the normal vectors will be
pointing into the region enclosed by the surface. So, if we walk along the edge of the surface, i.e. the
curve C, in the direction indicated with our head pointed into the region enclosed by the surface (i.e. in
the same direction as the normal vectors) then our left hand will be over the region. Therefore the
direction indicated in the sketch is the positive orientation of C.
If you have trouble visualizing the direction of the curve simply get a cup or bowl and put it on its side
with a piece of paper behind it. Sketch a set of axis on the piece of paper that will represent the plane the
cup/bowl is sitting in front of to really help with the visualization. Then cut out a little stick figure and
put a face on the “front” side of it and color the left hand a bright color so you can quickly see it. Now,
on the edge of the cup/bowl/whatever you place the stick figure with it’s head pointing in the direction of
the normal vectors (into the paraboloid/cup/bowl in our case) with its left hand over the surface. The
direction that the “face” on the stick figure is facing is the direction you’d need to walk along the surface
to get the positive orientation for C.
Step 2
   
∫∫
S
curl F  dS =
C
∫ dr
F
We’ve been given the vector field in the problem statement so we don’t need to worry about that. We
will need to deal with C.

Calculus II
In this case C is the curve we get by setting the two equations in the problem statement equal. Doing this
gives,
−2 = 6 − 4 y 2 − 4 z 2 → 4 y2 + 4z2 = 8 ⇒ y2 + z2 = 2
We will see following sketch of C if we are in front of the paraboloid and look directly along the x-axis.
One possible parameterization of C is given by,

r ( t ) = −2, 2 sin t , 2 cos t 0 ≤ t ≤ 2π
The x component of the parameterization is –2 because C lies at x = −2 .
Step 3
Since we know we’ll need to eventually do the line integral we know we’ll need the following dot
product.
  
F ( r ( t ) )r ' ( t ) = 2 cos 2 t − 1, 2 cos t + 4 2 sin 3 t , 6  0, 2 cos t , − 2 sin t
= 2 cos t ( )
2 cos t + 4 2 sin 3 t − 6 2 sin t
=
2 cos 2 t + 8cos t sin 3 t − 6 2 sin t
(1 − sin ( 2t ) ) + 8cos t sin 3 t − 6 2 sin t
=
Don’t forget to plug the parameterization of C into the vector field!
We also did a little simplification on the first term with an eye towards the integration.
Step 4
Okay, let’s go ahead and evaluate the integral using Stokes’ Theorem.

Calculus II
   
∫∫ curl F dS = ∫ F dr
S C
∫ (1 − sin ( 2t ) ) + 8cos t sin t − 6 2 sin t dt

2π
= 3
0
( )
2π
t + 12 cos ( 2t ) + 2sin 4 t + 6 2 sin t dt
= 2π
=
0
     
3. Use Stokes’ Theorem to evaluate ∫ F  dr where F =
− yz i + ( 4 y + 1) j + xy k and C is is the circle
C
of radius 3 at y = 4 and perpendicular to the y-axis. C has a clockwise rotation if you are looking down
the y-axis from the positive y-axis to the negative y-axis. See the figure below for a sketch of the curve.
Step 1
Okay, we are going to use Stokes’ Theorem in the following direction.
   
∫
C
F  dr = ∫∫
S
curl F dS

So, let’s first compute curl F since that is easy enough to compute and might be useful to have when we
go to determine the surface S we’re going to integrate over.
The curl of the vector field is then,

Calculus II
  
i j k
 ∂ ∂ ∂       
curl F = = xi − yj + zk − yj = xi − 2 yj + zk
∂x ∂y ∂z
− yz 4 y + 1 xy
Step 2
Now we need to find a surface S with an orientation that will have a boundary curve that is the curve
shown in the problem statement, including the correct orientation. This can seem to be a daunting task at
times but it’s not as bad as it might appear to be. First we know that the boundary curve needs to be a
circle. This means that we’re going to be looking for a surface whose cross section is a circle and we
know of several surfaces that meet this requirement. We know that spheres, cones and elliptic
paraboloids all have circles as cross sections.
The question becomes which of these surfaces would be best for us in this problem. To make this
decision remember that we’ll eventually need to plug this surface into the vector field and then take the
dot product of this with normal vector (which will also come from the surface of course).
In general it won’t be immediately clear from the curl of the vector field by itself which surface we
should use and that is the case here. The curl of the vector field has all three components and none of
them are that difficult to deal with but there isn’t anything that suggests one surface might be easier than
the other.
So, let’s consider a sphere first. The issue with spheres is that its parameterization and normal vector are
lengthy and many lead to messy integrands. So, because the curl of the vector field does have all three
components to it which may well lead to long and/or messy integrands we’ll not work with a sphere for
this problem.
Now let’s think about a cone. Equations of cones aren’t that bad but they will involve a square root and
=
in this case would need to be in the form y ax 2 + bz 2 because the boundary curve is centered on the
y-axis. The normal vector will also contain roots and this will often lead to messy integrands. So, let’s
not work with a cone either in this problem.
That leaves elliptic paraboloids and we probably should have considered them first. The equations are
simple and the normal vectors are even simpler so they seem like a good choice of surface for this
problem.
Note that we’re not saying that spheres and cones are never good choices for the surface. For some vector
fields the curl may end up being very simple with one of these surfaces and so they would be perfectly
good choices.
Step 3
We have two possibilities for elliptic paraboloids that we could use here. Both will be centered on the y-
axis but one will open in the negative y direction while the other will open in the positive y direction.
Here is a couple of sketches of possible elliptic paraboloids we could use here.

Calculus II
Let’s get an equation for each of these. Note that for each of these if we set the equation of the paraboloid
and the plane y = 4 equal we need to get the circle x 2 + z 2 = 9 since this is the boundary curve that
should occur at y = 4 .
Let’s get the equation of the first paraboloid (the one that opens in the negative y direction. We know that
the equation of this paraboloid should be y =a − x 2 − z 2 for some value of a. As noted if we set this
equal to y = 4 and do some simplification we know what equation we should get. So, let’s set the two
equations equal.
4 = a − x2 − z 2 → x2 + z 2 = a − 4 = 9 → a = 13
As shown we know that the a − 4 should be 9 and so we must have a = 13 . Therefore the equation of
the paraboloid that open in the negative y direction is,

Calculus II
y = 13 − x 2 − z 2
Next, let’s get the equation of the paraboloid that opens in the positive y direction. The equation of this
paraboloid will be in the form y = x 2 + y 2 + a for some a. Setting this equal to y = 4 gives,
4 =x 2 + z 2 + a → x 2 + z 2 =4 − a =9 → a =−5
The equation of the paraboloid that opens in the positive y direction is then,
y = x2 + z 2 − 5
Either of these surfaces could be used to do this problem.
Step 4
We now need to determine the orientation of the normal vectors that will induce a positive orientation of
the boundary curve, C, that matches the orientation that was given in the problem statement.
We’ll find the normal vectors for each surface despite the fact that we really only need to do it for one of
them since we only need one of the surfaces to do the problem as noted in the previous step. Determining
the orientation of the surface can be a little tricky for some folks so doing an extra one might help see
what’s going on here.
Remember that what we want to do here is think of ourselves as walking along the boundary curve of the
surface in the direction indicated while our left hand is over the surface itself. We now need to determine
if we are walking along the outside of the surface or the inside of the surface.
If we are walking along the outside of the surface then our heads, and hence the normal vectors, will be
pointing away from the region enclosed by the surface. On the other hand if we are walking along the
inside of the surface then our heads, and hence the normal vectors, will be pointing into the region
enclosed by the surface.
To help visualize this for our two surfaces it might help to get a cup or bowl that we can use to represent
the surface. The edge of the cup/bowl will then represent the boundary curve. Next cut out a stick figure
and put a face on one side so we know which direction we’ll be walking and brightly color the left hand
to make it really clear which side is the left side.
Now, put the cup/bowl on its side so it looks vaguely like the surface we’re working with and put the
stick figure on the edge with the face pointing in the direction the curve is moving and the left hand over
the cup/bowl. Do we need to put the stick figure on the inside or outside of the cup/bowl to do this?
Okay, let’s do this for the first surface, y = 13 − x 2 − z 2 . In this case our stick figure would need to be
standing on the inside of the cup/bowl/surface. Therefore the normal vectors on the surface would all
need to be point in towards the region enclosed by the surface. This also will mean that all the normal
vectors will need to have a negative y component. Again, to visualize this take the stick figure and move
it into the region and toward the end of cup/bowl/surface and you’ll see it start to point more and more in
the negative y direction (and hence will have a negative y component). Note that the x and z component
can be either positive or negative depending on just where we are on the interior of the surface.

Calculus II
Now, let’s take a look at the first surface, y = x 2 + z 2 − 5 . For this surface our stick figure would need to
be standing on the outside of the cup/bowl/surface. So, in this case, the normal vectors would point out
away from the region enclosed by the surface. These will also have a negative y component and you can
use the method we discussed in the above paragraph to help visualize this.
Step 5
We now need to start thinking about actually computing the integral. We’ll write the equation of the
surface as,
f ( x, y, z ) = 13 − x 2 − z 2 − y = 0
 ∇f −2 x, −1, −2 z
=n =
∇f ∇f
Note as well that this does have the correct orientation because the y component is negative.
  −2 x, −1, −2 z
curl F n = x, −2 (13 − x 2 − z 2 ) , z 
∇f
= 1
∇f ( −2 x 2
+ 2 (13 − x 2 − z 2 ) − 2 z 2 )
= 1
∇f ( 26 − 4 x 2
− 4z2 )
Step 6
Now, applying Stokes’ Theorem to the integral and converting to a “normal” double integral gives,
   
∫
C
F  dr = ∫∫ curl
S
F dS
= ∫∫ ( 26 − 4 x − 4 z 2 ) dS
1 2
∇f
S
= ∫∫ ( 26 − 4 x − 4 z 2 ) ∇f dA
1 2
∇f
D
= ∫∫ 26 − 4 x − 4 z 2 dA
2
Step 7
To finish this integral out then we’ll need to convert to polar coordinates using the following polar
coordinates.

Calculus II
= θ
In this case D is just the disk x 2 + z 2 ≤ 9 and so the limits for the integral are,
0 ≤ θ ≤ 2π
0≤r ≤3

   
∫
C
F  dr = ∫∫
S
curl F dS
∫ ∫ ( 26 − 4r ) r dr dθ
2π 3
= 2
0 0
π
2 3
= ∫ ∫ 26r − 4r dr dθ
3
0 0
2π
= ⌠ (13r 2 − r 4 ) dθ
3
⌡0 0
2π
= ∫ 36 dθ
0
= 72π
Don’t forget to pick up an extra r from converting the dA to polar coordinates.
     
4. Use Stokes’ Theorem to evaluate ∫ F  dr where F = ( 3 yx 2
+ z 3 ) i + y 2 j + 4 yx 2 k and C is is
C
triangle with vertices ( 0, 0,3) , ( 0, 2, 0 ) and ( 4, 0, 0 ) . C has a counter clockwise rotation if you are
above the triangle and looking down towards the xy-plane. See the figure below for a sketch of the curve.

Calculus II
Step 1
Okay, we are going to use Stokes’ Theorem in the following direction.
   
∫
C
F  dr = ∫∫S
curl F dS

So, let’s first compute curl F since that is easy enough to compute and might be useful to have when we
go to determine the surface S we’re going to integrate over.
The curl of the vector field is then,

  
i j k
 ∂ ∂ ∂       
curl F = = 4 x 2 i + 3 z 2 j − 3 x 2 k − 8 yxj = 4 x 2 i + ( 3 z 2 − 8 yx ) j − 3 x 2 k
∂x ∂y ∂z
3 yx 2 + z 3 y2 4 yx 2
Step 2
Now we need to find a surface S with an orientation that will have a boundary curve that is the curve
shown in the problem statement, including the correct orientation.
In this case we can see that the triangle looks like the portion of a plane and so it makes sense that we use
the equation of the plane containing the three vertices for the surface here.
The curl of the vector field looks a little messy so using a plane here might be the best bet from this
perspective as well. It will (hopefully) not make the curl of the vector field any messier and the normal
vector, which we’ll get from the equation of the plane, will be simple and so shouldn’t make the curl of
the vector field any worse.

Calculus II
Step 3
Determining the equation of the plane is pretty simple. We have three points on the plane, the vertices,
and so we can quickly determine the equation.
First, let’s “label” the points as follows,
=P ( 4,=
0, 0 ) Q (=
0, 2, 0 ) R ( 0, 0,3)
Then two vectors that must lie in the plane are,

 
QP = 4, −2, 0 QR = 0, −2,3
To write the equation of a plane recall that we need a normal vector to the plane. Now, we know that the
cross product of these two vectors will be orthogonal to both of the vectors. Also, since both of the
vectors lie in the plane the cross product will also be orthogonal, or normal, to the plane. In other words,
we can use the cross product of these two vectors as the normal vector to the plane.
The cross product is,

  
i j k
    
QP × QR = 4 −2 0 =−6i − 12 j − 8k
0 −2 3
Now we can use any of the points in our equation. We’ll use Q for our point. The equation of the plane
is then,
−6 ( x − 0 ) − 12 ( y − 2 ) − 8 ( z − 0 ) =0 → − 6 x − 12 y − 8 z =−24
3x + 6 y + 4 z =
12
Note that we divided the equation by –2 to make the equation a little “nicer” to work with.
Step 4
We now need to determine the orientation of the normal vectors that will induce a positive orientation of
the boundary curve, C, that matches the orientation that was given in the problem statement.
Remember that what we want to do here is think of ourselves as walking along the boundary curve of the
surface in the direction indicated while our left hand is over the plane. We now need to determine if we
are walking along the top or bottom of the plane.
If we are walking along the top of the plane then our heads, and hence the normal vectors, will be
pointing in a generally upwards direction. On the other hand if we are walking along the bottom of the
plane then our heads, and hence the normal vectors, will be pointing generally downwards.
To help visualize this for our plane it might help to cut out a triangular piece of paper that we can use to
represent the plane. The edge of the piece of paper will then represent the boundary curve. Next cut out a

Calculus II
stick figure and put a face on one side so we know which direction we’ll be walking and brightly color the
left hand to make it really clear which side is the left side.
Now, hold the piece of paper so that it looks vaguely like the surface we’re working with and put the stick
figure on the edge with the face pointing in the direction the curve is moving and the left hand over the
cup/bowl. Doing this we’ll quickly see that we must be walking along the top of the surface. Therefore
the normal vectors on the surface need to be pointing in a generally upwards direction (and hence will
have a positive z component).
Step 5
We now need to start thinking about actually computing the integral. We’ll write the equation of the
surface as,
z =3 − 34 x − 32 y
Recall that if we aren’t going to parameterize the surface we need it to be written as z = g ( x, y ) so that
the magnitude of the normal vector will eventually cancel.
Now, that we have the surface written in the “proper” form let’s define,
f ( x, y, z ) = z − 3 + 34 x + 32 y = 0
The unit normal vector for the surface is then,
 ∇f 3
, 32 ,1
=n = 4
∇f ∇f
Note as well that this does have the correct orientation because the z component is positive.
 
( )
3
, 32 ,1
4 x 2 , 3 ( 3 − 34 x − 32 y ) − 8 yx , −3 x 2 
2
curl=
F n 4
∇f
= 1
∇f ( 3x 2
+ 92 ( 3 − 34 x − 32 y ) − 12 xy − 3 x 2
2
)
= 1  9 ( 3 − 3 x − 3 y )2 − 12 xy 
∇f 2 4 2 
Don't forget to plug the equation of the surface into the curl of the vector field.
Step 6
Now, applying Stokes’ Theorem to the integral and converting to a “normal” double integral gives,

Calculus II
   
∫ F dr = ∫∫ curl F dS
C S
 9 ( 3 − 3 x − 3 y )2 − 12 xy  dS
= ∫∫
S
1
∇f 2 4 2 
 9 ( 3 − 3 x − 3 y )2 − 12 xy  ∇f dA
= ∫∫
D
1
∇f 2 4 2 
∫∫ ( 3 − x − 32 y ) − 12 xy dA
2
= 9
2
3
4
D
Step 7
To finish this integral we just need to determine D. In this case D just the triangle in the xy-plane that lies
below the plane. Here is a quick sketch of D.
The integral doesn’t seem to suggest one integration order over the other so let’s use the following set of
limits for our integral.
0≤ x≤4
0 ≤ y ≤ 2 − 12 x

Calculus II
   
∫ F dr = ∫∫ curl F dS
C S
4 2 − 12 x
= ⌠ ∫ ( 3 − 34 x − 32 y )
2
9
− 12 xy dy dx
⌡0 0 2
( )
4 2 − 12 x
= ⌠
 − ( 3 − 34 x − 32 y ) − 6 xy
3 2
dx
⌡0 0
= ∫ ( 3 − 34 x ) − 6 x ( 2 − 12 x ) dx
4 3 2
∫ (3 − x )
4 3
= 3
4 − 24 x + 12 x 2 − 32 x 3 dx
0
( )
4
=− 13 ( 3 − 34 x ) − 12 x 2 + 4 x 3 − 83 x 4
4
= −5
Divergence Theorem
     
1. Use the Divergence Theorem to evaluate ∫∫ F  dS where F = yx 2
i + ( xy 2
− 3 z 4
) j + ( x 3
+ y 2
) k
S
and S is the surface of the sphere of radius 4 with z ≤ 0 and y ≤ 0 . Note that all three surfaces of this
solid are included in S.
Step 1

Calculus II
surface.
Note as well here that because we are including all three surfaces shown above that the surface does
enclose (or is the boundary curve if you want to use that terminology) the portion of the sphere shown
above.
Step 2
We are going to use the Divergence Theorem in the following direction.

Calculus II
  
∫∫
S
F  dS = ∫∫∫E
div F dV
where E is just the solid shown in the sketches from Step 1.
Because E is a portion of a sphere we’ll be wanting to use spherical coordinates for the integration. Here
are the spherical limits we’ll need to use for this region.
π ≤ θ ≤ 2π
2 ≤ϕ ≤π
π
0≤ρ ≤4
One of the restrictions on the region in the problem statement was y ≤ 0 . This means that if we look at
this from above we’d see the portion of the circle of radius 4 that is below the x axis and so we need the
given range of θ above to cover this region.
We were also told in the problem statement that z ≤ 0 and so we only want the portion of the sphere that
is below the xy-plane. We therefore need the given range of ϕ to make sure we are only below the xy-
plane.
We’ll also need the divergence of the vector field so here is that.
 ∂ ∂ ∂
div=
F
∂x
( yx 2 ) + ( xy 2 − 3 z 4 ) + ( x3 + y=
∂y ∂z
2
) 4 xy
Step 3
Now let’s apply the Divergence Theorem to the integral and get it converted to spherical coordinates
while we’re at it.
  
∫∫
S
F  dS = ∫∫∫
E
div F dV
= ∫∫∫ 4 xy dV
E
2π
∫ π ∫ 4 ( ρ sin ϕ cos θ )( ρ sin ϕ sin θ ) ( ρ sin ϕ ) d ρ dϕ dθ

π
=⌠
4
 2
⌡π 1
2
0
2π π
=⌠
4

⌡π ∫π∫
1
2
0
4 ρ 4 sin 3 ϕ cos θ sin θ d ρ dϕ dθ
Don’t forget to pick up the ρ 2 sin ϕ when converting the dV to spherical coordinates.
Step 4
All we need to do then in evaluate the integral.

Calculus II
  2π π
⌠ 4
∫∫S F  dS =
⌡π ∫12 π ∫0
 4 ρ 4 sin 3 ϕ cos θ sin θ d ρ dϕ dθ
2π π
=⌠ ⌠ ( 4 ρ 5 sin 3 ϕ cos θ sin θ ) dϕ dθ
4
 ⌡
⌡π 12 π
5
0
2π
sin ϕ (1 − cos 2 ϕ ) cos θ sin θ dϕ dθ
π
= ⌠
 ∫π 4096
⌡π 1
2
5
2π
( )
π
5 ( )
⌠ − 4096 cos ϕ − 1 cos3 ϕ cos θ sin θ
= dθ

⌡π 3 1π
2
2π
=∫ 8192
15 cos θ sin θ dθ
π
2π
=∫ 4096
15 sin ( 2θ ) dθ
π
2π
= cos ( 2θ ) π
=
2048
15 0
Make sure you can do use your trig formulas as we did here to deal with these kinds of integrals!
     
2. Use the Divergence Theorem to evaluate ∫∫ F dS =
where ( )
F sin (π x ) i + zy 3 j + z 2 + 4 x k and S
S
is the surface of the box with −1 ≤ x ≤ 2 , 0 ≤ y ≤ 1 and 1 ≤ z ≤ 4 . Note that all six sides of the box are
included in S.
Step 1

Calculus II
surface.
Note as well here that because we are including all six sides of the box shown above that the surface does
enclose (or is the boundary curve if you want to use that terminology) for the box.
Step 2
We are going to use the Divergence Theorem in the following direction.
  
∫∫
S
F  dS = ∫∫∫
E
div F dV
E is just a box and the limits defining it where given in the problem statement. The limits for our integral
will then be,

Calculus II
−1 ≤ x ≤ 2
0 ≤ y ≤1
1≤ z ≤ 4
 ∂ ∂ ∂
=
div F ( sin (π x ) ) + ( zy 3 ) + ( z 2 +=
4 x ) π cos (π x ) + 3 zy 2 + 2 z
∂x ∂y ∂z
Step 3
Now let’s apply the Divergence Theorem to the integral and get it converted to a triple integral.
  
∫∫ F
S
 dS = ∫∫∫ div
E
F dV
= ∫∫∫ π cos (π x ) + 3zy + 2 z dV

2
π cos (π x ) + 3zy 2 + 2 z dz dy dx
2 1 4
= ∫ ∫∫−1 0 1
Step 4
 
π cos (π x ) + 3zy 2 + 2 z dz dy dx
2 1 4
∫∫ dS
= F
S
∫ ∫∫
−1 0 1
 ⌠ (π z cos (π x ) + 32 z 2 y 2 + z 2 ) dy dx
2
= ⌠
1 4
⌡−1 ⌡0 1
∫ ∫ 3π cos (π x ) +
2 1
= 45
2 y 2 + 15 dy dx
−1 0
= ⌠ ( 3 yπ cos (π x ) + 152 y 3 + 15 y ) dx
2 1
⌡−1 0
∫ 3π cos (π x ) +
2
= 45
2 dx
−1
( 3sin (π x ) + x )
2
= 45
2 = 135
2
−1π
     
3. Use the Divergence Theorem to evaluate ∫∫ F  dS where =
F 2 xzi + (1 − 4 xy 2
) j + ( 2 z − z 2
) k and
S
6 2 x 2 − 2 y 2 and the plane z = 0 . Note that both of the

S is the surface of the solid bounded by z =−
surfaces of this solid included in S.
Step 1

Calculus II
surface. The bottom “cap” of the elliptic paraboloid is also included in the surface but isn’t shown.
Note as well here that because we are including both of the surfaces shown above that the surface does
enclose (or is the boundary curve if you want to use that terminology) the region.
Step 2
  
∫∫ F dS = ∫∫∫ div F dV
S E
The region D for that we’ll need in converting the triple integral into iterated integrals is just the
intersection of the two surfaces from the problem statement. This is,
0 =6 − 2 x 2 − 2 y 2 → x 2 + y 2 =3
So, D is a disk and so we’ll eventually be doing cylindrical coordinates for this integral. Here are the
cylindrical limits for the region E.
0 ≤ θ ≤ 2π
0≤r ≤ 3
0 ≤ z ≤ 6 − 2 x 2 − 2 y 2 = 6 − 2r 2
Don’t forget to convert the z limits into cylindrical coordinates as well!

Calculus II
 ∂ ∂ ∂
div F = ( 2 xz ) + (1 − 4 xy 2 ) + ( 2 z − z 2 ) =−
2 8 xy
∂x ∂y ∂z
Step 3
Now let’s apply the Divergence Theorem to the integral and get it converted to cylindrical coordinates
while we’re at it.
  
∫∫
S
F  dS = ∫∫∫ div F
E
dV
= ∫∫∫ 2 − 8 xy dV
E
2π
= ⌠ ( 2 − 8r 2 cos θ sin θ ) r dz dr dθ
3 6− 2 r 2
 ⌠
⌡0 ⌡0 ∫0
2π
= ⌠
3 6− 2 r 2
 ⌠
⌡0 ⌡0 ∫0
2r − 8r 3 cos θ sin θ dz dr dθ
Don’t forget to pick up the r when converting the dV to cylindrical coordinates.
Step 4
  ⌠ 2π 3 6 − 2 r 2
∫∫S F dS ⌡0 ⌠⌡0 ∫0 2r − 8r cos θ sin θ dz dr dθ
= 3
2π
⌠
( 2r − 8r 3 cos θ sin θ ) z
3 6− 2 r 2
=  ⌠ dr dθ
⌡0 ⌡0 0
2π
⌡0 0 (
⌠ 2r − 8r 3 cos θ sin θ )( 6 − 2r 2 ) dr dθ
3
= ∫
2π
= ⌠ 12r − 4r 3 − ( 48r 3 − 16r 5 ) cos θ sin θ dr dθ
3
⌡0 ∫ 0
2π
= ⌠ ( 6r 2 − r 4 − (12r 4 − 83 r 6 ) cos θ sin θ )
3
 dθ
⌡0 0
2π
= ∫ 9 − 36 cos θ sin θ dθ
0
π
= ∫ 9 − 18sin ( 2θ ) dθ
2
( 9θ − 9 cos ( 2θ ) ) =
π 2
= 18π
0

Calculus II

CalcIII SurfaceIntegrals Solutions

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CALCULUS III

Solutions to Practice Problems

© 2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

© 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

1. Write down a set of parametric equations for the plane 7 x + 3 y + 4 z =

For this problem let’s solve for z to get,

2. Write down a set of parametric equations for the plane 7 x + 3 y + 4 z =

© 2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx

The parametric equation for this plane is then,

Here is a sketch of this region.

© 2007 Paul Dawkins 4 http://tutorial.math.lamar.edu/terms.aspx

The conversion equations are,

Doing that gives,

© 2007 Paul Dawkins 5 http://tutorial.math.lamar.edu/terms.aspx

4. The portion of y =4 − x 2 − z 2 that is in front of y = −6 .

5. The portion of the sphere of radius 6 with x ≥ 0 .

© 2007 Paul Dawkins 6 http://tutorial.math.lamar.edu/terms.aspx

We can restrict x to this range if we restrict θ to the range − 12 π ≤ θ ≤ 12 π .

© 2007 Paul Dawkins 7 http://tutorial.math.lamar.edu/terms.aspx

Here are the equations we get if we do that.

© 2007 Paul Dawkins 8 http://tutorial.math.lamar.edu/terms.aspx

9 that is inside the cylinder x 2 + y 2 =

The parameterization for the full plane is then,

Now, we what we really need is,

© 2007 Paul Dawkins 9 http://tutorial.math.lamar.edu/terms.aspx

So, the surface area is simply,

The parameterization for the full sphere is,

Now, we what we really need is,

© 2007 Paul Dawkins 10 http://tutorial.math.lamar.edu/terms.aspx

= 625sin 4 ϕ ( cos 2 θ + sin 2 θ ) + 625sin 2 ϕ cos 2 ϕ

= 625sin 2 ϕ ( sin 2 ϕ + cos 2 ϕ )

© 2007 Paul Dawkins 11 http://tutorial.math.lamar.edu/terms.aspx

Now, we what we really need is,

© 2007 Paul Dawkins 12 http://tutorial.math.lamar.edu/terms.aspx

Now, we what we really need is,

Where D is the disk u 2 + v 2 ≤ 4 .

© 2007 Paul Dawkins 13 http://tutorial.math.lamar.edu/terms.aspx

The polar coordinate limits for this D is,

So, the integral to converting to polar coordinates gives,

∫∫ z + 3 y − x dS where S is the portion of z =2 − 3 y + x 2 that lies over the triangle in the

xy-plane with vertices ( 0, 0 ) , ( 2, 0 ) and ( 2, −4 ) .

© 2007 Paul Dawkins 14 http://tutorial.math.lamar.edu/terms.aspx

© 2007 Paul Dawkins 15 http://tutorial.math.lamar.edu/terms.aspx

We could use either of the following sets of limits to describe D.

so we’ll use the following formula for the surface integral.

The integral is then,

© 2007 Paul Dawkins 16 http://tutorial.math.lamar.edu/terms.aspx

© 2007 Paul Dawkins 17 http://tutorial.math.lamar.edu/terms.aspx

So, D will be the disk x 2 + z 2 ≤ 2 .

so we’ll use the following formula for the surface integral.

The integral is then,

© 2007 Paul Dawkins 18 http://tutorial.math.lamar.edu/terms.aspx

Converting the integral to polar coordinates gives,

∫∫S 40 y dS ⌠⌡0 120r 2 36r 2 + 1 ( r ) dr dθ

We’ll use the following substitution to do the integral.

So, using the substitution the integral becomes,

© 2007 Paul Dawkins 19 http://tutorial.math.lamar.edu/terms.aspx

We can now easily finish evaluating the integral.

© 2007 Paul Dawkins 20 http://tutorial.math.lamar.edu/terms.aspx

where u and v will be chosen as needed when doing the parameterization.