Math 203: Spring 2022 FINAL EXAM

Duration: 120 minutes

I pledge my honor that I did not receive nor give any help during
this exam and I am aware of the University’s honor code.

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Check your section

 Section 1 (Mon. & Wed.)  Section 2 (Mon. & Wed.) Attila Aşkar

 Section 3 (Tue. & Thu.)  Section 4 (Tue. & Thu.) Varga Kalantarov

Show your work by writing what you do, write clearly the formulas
you use.

Qu 1: /20 Qu 2: /20 Qu 3: /20 Qu 4: /20 Qu 5: /20 Qu 6: /20 SUM: /120

Notation

You may use r or x for position vector as your choice, or

an arrow for denoting any vector.

You may use |r| or ||x|| for the length (Norm) of a vector
as you choose.
 B B 
2
r dr dθ dz r sin φ dr dφ dθ F · dr = U  F · dr = 0

A A C
p
q Ux2 + Uy2 + Uz2 ∇U
fx2 + fy2 + 1 dA dA (−fx , −fy , 1) dA dA
|Uz | Uz
1. Given the planes A : x + y + z = 1, B : x − 2y + 3z = −5
(i) Find the angle between the planes A and B,

(ii) Find the equation of the line L of intersection of A and B.

Solution

(i) The normal vector of A is the vector n1 = i + j + k and n2 = i − 2 j + 3 k is the

normal vector of B. Thus
n1 · n2 2 2
cos θ = =√ √ Hence: θ = arccos √
| n1 || n2 | 3 14 42
(ii) First we find a point on the line of intersection of these planes. To this end we take
z = 0 and find solve the system

x+y =1 x − 2y = −5

This sytem has a unique solution x = −1, y = 2. So the point (−1, 2, 0) is on the line of
intersection of A and B. It remains to find a vector v that is parallel to the line L.

v = n1 × n2 = 5i − 2j − 3k

Thus the line L of intersection of the planes A and B has the following parametric equations

x = −1 + 5t, y = 2 − 2t, z = −3t.

For (ii) there is also a second way: select one of the coordinates as s, a parameter not necessarily equal to t, and
calculate the other coordinates in terms of s.

For instance, set z = s, then: x+ y = 1-3s x - 2y = - 5 -3s

Solving for x and y from the two equations in terms of s gives: x = - 1 - 5s/3 y = 2s/3 + 2.

The change of variables s = -3t, yields the same result.

2. Verify whether the following functions are continuous at the point (0, 0):
 
 2xy 2 (x, y) 6= (0, 0),  2x2 y 2 , (x, y) 6= (0, 0),
x +y
(i) f (x, y) = (ii) g(x, y) = x +y
1, (x, y) = (0, 0), 0, (x, y) = (0, 0),

(i) Approaching the point (0, 0) along the line y = kx we get

kx2 k
f (x, kx) = 2 2 2
=
x +k x 1 + k2
Limit does not exist because it is not a unique value. Thus the function is not continuous at
the point (0, 0).

(ii) It is clear that

x2
|g(x, y)| = |x| 2 ≤ |x| for all (x, y) 6= (0, 0).
x + y2
Hence
g(x, y) → 0 as (x, y) → (0, 0).

Thus the function is continuous at the point (0, 0).

Consider the function
x2 y
g(x, y) = .
x4 + y 2
This function tends to zero along each line y = kx:

kx3 kx
lim g(x, kx) = lim = lim =0
x→0 x→0 x4 + k 2 x2 x→0 x2 + k 2

But along the lines y = kx2 we have

kx4 k
lim g(x, kx2 ) = lim =
x→0 x→0 x4 + k 2 x4 1 + k2
Soalong these lines the function tends to different numbers. Hence the limit does not exist.
3. Find and classify all critical points of the following functions.

x3 − y 3
(i) F (x, y) = xy + (ii) h(x, y) = x3 + y 3
3

x3 − y 3
SOLUTION. (i) F (x, y) = xy + → F x = y + x2 = 0 Fy = x − y 2 = 0
3
Substituting y = −x2 from Fx = 0 into the Fy = 0 yields:

x − x4 = 0 → x(1 − x3 ) = x(1 − x)(1 + x + x2 ) = 0 → x1 = 0 x2 = 1

The corresponding values for y are: y1 = 0 y2 = −1.

For the types of the critical points, we need: Fxx = 2x Fxy = Fyx = 1 Fyy = −2y →
   
F F  2x 1 
 xx xy 
Hessian =  =  = −(4xy + 1) →
 
Fyx Fyy   1 −2y 
 
Hessian = −1 < 0 Hessian =3>0
 
(0,0) (1,−1)

Conclusion: At (0, 0) there is a saddle point; At (1, −1) there is a Minimum point.

(ii) h(x, y) = x3 + y 3 → hx = 3x2 = 0 hy = 3y 2 − 0 → x = y = 0.

So the point (0, 0) is a critical point. For the types of of the critical points, we have: hxx =
6x hxy = hyx = 0 hyy = 6y →
   
h h  6x 0  
 xx xy  
Hessian =  =  = 36xy → Hessian =0
 
hyx hyy   0 6y  (0,0)

This test is inconclusive.

But for each points (−, 0), (0, ) with  > 0 we have

h(−, 0) = −3 < h(0, 0) = 0, h(, 0) = 3 > h(0, 0) = 0.

Thus the point (0, 0) is the sadle point.

-3 pts who did not write last 3 rows
4. For U (x, y, z) = e2y sin(2x) + y 2 + z 2 , calculate

(i) div (∇U ) and (ii) curl (∇U ),

SOLUTION.
∂ 2U ∂ 2U ∂ 2U
(i) div (∇U ) ≡ ∇2 U = + + → For U = e2y sin(2x) + y 2 + z 2 :
∂x2 ∂62 ∂z 2
div (∇U ) = 2 + 2 = 4

 
 i j k
 
 
∂
(ii) curl (∇U ) ≡  ∂x ∂ ∂ 
∂y ∂z 
 ∂U ∂U ∂U 
 ∂x ∂y ∂z
 ∂ 2U ∂ 2U   ∂ 2U ∂ 2U   ∂ 2U ∂ 2U 
=i − −j − + − k
∂y∂z ∂z∂y ∂x∂z ∂z∂x ∂x∂y ∂y∂x
For any U
curl (∇U ) = i0 + j0 + k0 = 0

Thus, without any need to take derivatives: curl (∇(e2y sin(2x) + y 2 + z 2 )) = 0

5. State the Green’s Theorem and verify it for the vector field

F = 2xyi + (x + y)j,

(i) on the disk D := {(x, y) : x2 + y 2 ≤ 1} and

(iia) on the trangle with vertices (0, 0), (1, 0), (1, 2).
SOLUTION.

Green’s Theorem If the components of the vector vector field F(x, y) = u(x, y)i + v(x, y)j are
continuojs in a simply connected domain D with piecewise smooth boundary C, then
 
(vx (x, y) − uy (x, y))dxdy = udx + vdy.
D C

(i) Let F = (u, v) = (2xy, (x + y)) and D := {(x, y) : x2 + y 2 ≤ 1} .

   1 2π
A := (vx (x, y) − uy (x, y))dxdy = (1 − 2x) dxdy = (1 − 2 cos t) rdθdr
D D 0
 0
1  2π
= rdr (1 − cos θ)dθ = π.
0 0

  
I= 2xy dx + (x + y) dy = (vx − uy ) dxdy = (1 − 2x) dxdy
C
 1 2π  1  2π
= (1 − 2 cos t) rdθdr = rdr (1 − cos θ)dθ = π
0 0 0 0

So for for the disk D, A = π → I=π

(ii) For the trangle with vertices (0, 0), (1, 0), (1, 2) :
  1 2x
((x + y)x − (2xy)y ) dA = (1 − 2x)dydx
D 0 0

  1 2x
((x + y)x − (2xy)y ) dA = (1 − 2x) dydx
 1
Ω 0 0
 1
2 2 4 3 1
= (2x − 4x )dx = x − x =− .
0 3 0 3
 The positively oriented boundary C of the triangel Ω is piecewise somooth and C = C1 ∪
C2 ∪ C3 , where

C1 = {(x, 0) : 0 ≤ x ≤ 1}, C2 = {(1, y) : 0 ≤ y ≤ 2}, C3 = {(1 − x, 2(1 − x)) : 0 ≤ x ≤ 1}

 
F · dr = (2xydx + (x + y)dy = I1 + I2 − I3 ,
C C
where
  
I1 = (2xydx + (x + y)dy, I2 = (2xydx + (x + y)dy, I3 = (2xydx + (x + y)dy
C1 C2 C3

Since y = 0 and dy = 0 on C1 we have I1 = 0.

 2
I2 = (1 + y)dy = 4
0
 1
13
4x2 + 6x dx = .


I3 =
0 3
Hence 
13 1
F · dr = 4 − =− .
C 3 3

6. Given the vector field

F = (z − y)i + xj − xk

and
(i) State the Stokes’ Theorem and use it to calculate the circulation of the vector field F
along the positively oriented (oriented counterclockwise) boundary C = {(x, y) : x2 + y 2 = 4}
of the semisphere x2 + y 2 + z 2 = 4, z ≥ 0.
(ii) Calculate the flux of F across the boundary of the semiball x2 + y 2 + z 2 ≤ 4, z ≥ 0.
SOLUTION (i). The surface is an open surface. Therefore we can apply the Stokes
Theorem for any surface that uses C as its boundary. In particular, the disk in the xy plane
that is enclosed by C or calculate directly. So, we have three expressions that give the same
result. We will choose the one that gives the result in the easiest way.
 
 i j k
 
 
∇ × F =  ∂x ∂ ∂ ∂  = (0, 0, 2) ≡ 2k
∂y ∂z 
(z − y) x −x
 


On C, z = 0 → F = −y i + xj = (−y, x)

z=0

By Stokes Theorem:
 With the ∇ × F calculated above, although it is simple, the calculation over the hemisphere
is rather detailed. The integral over the disk in the xy plane is the simplest:
  
I= (∇ × F ) · k dA = (2k) · k dA = 2 dA = 2(22 )π = 8π
A A A

Second easy method: we can also calculate the integral along C:

 
I= F · dr = (−y dx + x dy)
C C

Using the polar coordinates:

x = 2 cos θ, y = −2 sin θ → −y dx + x dy = 4 sin2 θ + 4 cos2 θ) dθ = 4dθ →

  θ=2π
I=4 dθ = 2 dθ = 4 · (2π) = 8π
C θ=0

SOLUTION (ii). Because the surface is claosed hemisphere with the bottom at z = 0,
we can use the divergence theorem.
 
F · n dS = ∇ · F dV

For the given vector field F = (z − y) i + xj − x k → ∇·F=0

We use our judgement to select the easiest integral. Thus:

 
I= ∇ · F dV = 0 dV = 0
V V