Section 6-2 : 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,

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
component of the vector x, y, 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.
 
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).

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
component of the vector x, y, z .

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,


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
 
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
−12i + 36 j − 18k
=

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 − 32 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

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,


r ( x, =
y) 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.
  
1, 0, − 13
rx = 0,1, − 12
ry =

  
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,

∫∫ dA = Area of D
D

So, the surface area is simply,

π =
( )
7 
2
=A ∫∫=
6 dA
7 7
∫∫ dA
= 7
[ Area =
of D ] 7 49
π
D
6
D
6 6
  6

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 12 π ≤ ϕ ≤ π .

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
=

Now, we what we really need is,

 
( −25sin ϕ cos θ ) + ( −25sin 2 ϕ sin θ ) + ( −25sin ϕ cos ϕ )
2 2 2 2
rθ × rϕ =

= 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
The integral for the surface area is then,
 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 −25cos ϕ 1
2
π
d θ ∫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-
4
plane bounded by y = x 5 , x = 1 and the x-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.

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

Now, we what we really need is,

 
rx × ry = ( − x ) + ( −2 ) + (1)
3 2 2 2
= x6 + 5

Step 3
The integral for the surface area is then,

x 1 5

A= ∫∫ x 6 + 5 dA= ⌠ ∫ x 6 + 5 dy dx
D
⌡0 0

As noted in the integral above D is just the ranges of x and y we found in Step 1.

Step 4
Now we just need to evaluate the integral to get the surface area.

1 x 1 5

x + 5 dy dx = ⌠
5
x
A= ⌠ ∫ 6 6
 y x + 5 dx
⌡0 0 ⌡0 0

(6 )
3 1

(x + 5)
1 3 3

∫
5 6 6
= x x + 5 dx = 1
9
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

Now, we what we really need is,

 
( −2v ) + ( 4 ) + ( 2u )=
2 2 2
ru × rv= 4u 2 + 4v 2 + 16= 2 u 2 + v 2 + 4

Step 3
The integral for the surface area is then,

=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 =

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
=A ∫∫ 2D
u 2 + v 2 + 4=
dA ∫ ∫
0 0
2r r 2 + 4 dr dθ

Step 5
Now we just need to evaluate the integral to get the surface area.

2π
3 2
⌠
(r + 4)
2π 2
∫ ∫ 2r r + 4 dr dθ =  dθ
2 2
A= 2
3
2

0 0
⌡0 0

( ) ( )
2π

( )
2π 3 3 3
= ⌠ 2
8 2 − 4 2 dθ= 2
82 − 8 θ = 32
π 8 − 1= 61.2712
⌡0 3 3
0
3