Download Problem 1. (10 pts.) Show that there is no function f : R3 → R for

Problem 1. (10 pts.) Show that there is no function f : R3 → R for which


2x − 5z

3x
∇f = 
3
y −x
Solution: Assume there such an f existed. Then, on one hand, by the properties of the
gradient
∇ × ∇f = 0.
On the other hand

 



→
→
→
3y 2
2x − 5z
j
i
k

 = det 
3x
̸ 0.
∇×

∂x
∂y
∂z  =  −4  =
3
y3 − x
3
2x − 5z 3x y − x
→
→
→
We got a contradition, hence the vector field (2x−5z) i +3x j +(y 3 −x) k is not conservative
(so such an f does not exist).
Problem 2. (10 pts.) Let C be the ellipse x2 /9 + y 2 /4 = 1 oriented counterclockwise and
[
]
→
let F = 2x + 3y, x + 4y .
∮
→ →
a) Find F ◦ T ds.
C
Solution: (a) Let R be the inside of the ellipse: R = {x2 /9 + y 2 /4 ≤ 1}. By Green’s
theorem
∮
∫∫
→ →
(∂x (x + 4y) − ∂y (2x + 3y)) dA =
F ◦ T ds =
C
R
∫∫
=
(−2) dA = −2 Area(R) = −2 · 3 · 2π = −12π.
∮
R
→
→
→
F ◦ n ds, where n is the normal vector pointing out.
b) Find
C
Solution: By the divergence theorem
∫∫
∮
→ →
∇ ◦ F dA =
F ◦ n ds =
R
C
∫∫
=
6 dA = 6 Area(R) = 36π.
R
c) Both integrals in part a) and in part b) are different from zero. Which one of the two
→
results implies that the vector field F is not conservative? Explain.
1
→
→
→
Solution: If F were conservative then the integral of F ◦ T along any closed curve would
be zero, so it is the part a) that gives the contradiction.
Problem 3. (10 pts.)
a=
5
b=
3
b
Let D be any elliptical disk whose center is at (−2, 1)
and whose axis are 3 and 5. (Thus the area of D is 15π).
Let C be the boundary of D.
Compute:
∮ (
)
(
)
3
ex + 3y 2 dx + 2x + 4y − y 2 dy
7
6
5
4
3
2
(−2,
1 1)
−7 −6 −5 −4 −3 −2 −1
−1
C
1 2 3 4
−2
−3
−4
−5
Solution: Using Green’s theorem we get that
∮
∫∫
∫∫
x3
2
2
2
x3
2
(e +3y )dx+(2x+4y−y )dy =
(∂x (2x+4y−y )−∂y (e +3y ))dA =
(2−3y)dA.
C
D
D
Let’s split the last integral into two
∫∫
∫∫
(2 − 6y) dA = 2
D
D
∫∫
dA − 6
y dA
D
and use some geometric information to avoid explicitly calculating them
∫∫
dA = Area(D) = 15π
D
∫∫
and
y dA = Area(D) · ȳ = 15π · 1 = 15π.
D
(we know that the centroid of an ellipse is at its center, hence ȳ = 1). Therefore,
∮
3
(ex + 3y 2 ) dx + (2x + 4y − y 2 ) dy = 2 · 15π − 6 · 15π = −60π.
C
Problem 4. (10 pts.) Recall that we learned about the following coordinates in R3 :


 x = r cos θ
 x = ρ sin ϕ cos θ
y = r sin θ
y = ρ sin ϕ sin θ
cylindrical:
spherical:


z=z
z = ρ cos ϕ
2
(a) Write the cylindrical coordinates z, r and θ in terms of spherical ones ρ, ϕ and θ.
Solution: By geometric construction we know that the coordinate θ has the same definition in cylindrical and spherical coordinates. The coordinate z has the same definition in
cylindrical and spherical coordinates, so we may use the formula coming from writing the
rectangular coordinates in terms of the spherical ones: z = ρ sin ϕ. Finally, to find r let’s inspect the expression for x in cylindrical and spherical coordinates. On one hand x = r cos θ,
on the other hand x = ρ sin ϕ cos θ, so r cos θ = ρ sin ϕ cos θ, i.e. r = ρ sin ϕ.

 z = ρ cos ϕ
r = ρ sin ϕ

θ = θ.
Answer:
∫∫∫
(b) Convert
f (r) dz dr dθ into the spherical coordinates.
V
Assume that W is the solid V in spherical coordinates, so worry only about the Jacobian.
Solution: Solution 1 (a.k.a. “usual way”):
Using the change of variables formula we get that
∫∫∫
∫∫∫
∂(z, r, θ) dρ dϕ dθ.
f (r) dz dr dθ =
f (ρ sin ϕ) ∂(ρ,
ϕθ)
V
W
Using the result of part (a) we get that

∂z
∂ρ
∂z
∂ϕ
∂z
∂θ
∂(z, r, θ)
= det 
∂(ρ, ϕ, θ)
∂r
∂ρ
∂r
∂ϕ
∂r
∂θ
∂θ
∂ρ
∂θ
∂ϕ
∂θ
∂θ

=

cos ϕ
sin ϕ 0
= det  −ρ sin ϕ ρ cos ϕ 0  = ρ.
0
0
1

Hence,
∫∫∫
∫∫∫
f (r) dz dr dθ =
f (ρ sin ϕ)ρ dρ dϕ dθ.
V
W
Solution 2 (a.k.a. “how to do this problem using what we already know about converting
integrals into cylindrical and spherical coordinates and not calculate any Jacobians”):
For a function g(x, y, z) we know that
∫∫∫
∫∫∫
g(x, y, z) dx dy dz =
g(r cos θ, r sin θ, z) r dz dr dθ
U
V
3
(1)
(where U is the solid V in rectangular coordinates) and
∫∫∫
∫∫∫
g(x, y, z) dx dy dz =
g(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ sin ϕ) ρ2 sin ϕ dρ dϕ dθ.
U
(2)
W
√
√
2 + y 2 )/
If
we
let
g(x,
y,
z)
=
f
(
x
x2 + y 2 = f (r)/r then the last integral in (1) is exactly
∫∫∫
f (r) dz dr dθ. For this choice of g the last integral in (2) is
V
∫∫∫
∫∫∫
f (ρ sin ϕ) 2
ρ sin ϕ dρ dϕ dθ =
f (ρ sin ϕ)ρ dρ dϕ dθ
ρ sin ϕ
W
W
which is exactly what we want.
[
]
→
Problem 5. (10 pts.) Let r = x, y be the position vector in R2 .
Show that ∇ ◦ (∇rα ) = 0 only for α = 0.
Hint: Use the product rules and chain rules for ∇ to compute ∇ ◦ (∇rα ) in terms of α.
Solution: First of all, we compute ∇(r2 ) = ∇(x2 + y 2 ) = 2⃗r, and by chain rule, ∇(r2 ) =
2r∇r. Since these two are the same, we may find the identity
⃗r
∇r = .
r
Now, from the chain rule we have
∇(rα ) = αrα−1 ∇r = αrα−2⃗r.
Using product rule ∇(f F ) = ∇f ◦ F + f ∇ ◦ F , we have
∇ ◦ ∇rα = ∇ ◦ (αrα−2⃗r)
(
)
= α ∇(rα−2 ) ◦ ⃗r + rα−2 ∇ ◦ ⃗r
(
)
= α (α − 2)rα−3 ∇r ◦ ⃗r + 2rα−2
(
)
r ◦ ⃗r
α−3 ⃗
α−2
= α (α − 2)r
+ 2r
r
(
)
= α (α − 2)rα−2 + 2rα−2 = α2 rα
which is the zero vector field if and only if α = 0.
Problem 6. (10 pts.) Let C be the y > 0 part of the circle x2 + y 2 = 1 oriented from
(−1, 0) to (1, 0).
Compute
∫
)
(√
( 2
)
1 + y 3 + 2x dy
3x + y dx +
C
Hint: Complete the path C with a segment on x− axis to a closed path and use Green’s
Theorem.
4
Solution: Let C1 be the straight line path on the x-axis from (−1, 0) to (1, 0), and D be
the region x2 + y 2 ≤ 1, y ≥
√ 0. Then ∂D = C1 − C considering the orientation. For brevity,
let P = 3x2 + y and Q = 1 + y 2 + 2x. Using Green’s theorem on D, we have
∫∫
∫
∂Q ∂P
−
dA =
P dx + Qdy.
∂y
D ∂x
C1 −C
For the double integral, we have
∫∫
∫∫
∂Q ∂P
−
dA =
2 − 1dA = Area(D) = π/2
∂y
D
D ∂x
and for C1 , we use parametrization x = t, y = 0 on −1 ≤ t ≤ 1 to have
∫
∫ 1
P dx + Qdy =
(3t2 + 0)dt = 2.
−1
C1
Thus
∫
∫
∫∫
P dx + Qdy −
P dx + Qdy =
C
C1
D
∂Q ∂P
−
dA
∂x
∂y
π
=2− .
2
Problem 7. (10 pts.) Find the surface area of the part of the surface z = y 2 − x2 that is
inside the cylinder x2 + y 2 = 4.
Solution: The surface lying inside x2 + y 2 = 4 indicates x2 + y 2 ≤ 4. Using the parametrization of a graph z = y 2 − x2 , we have
α(u, v) = (u, v, v 2 − u2 ),
(u, v) ∈ D
where D = {(u, v) : u2 + v 2 ≤ 4}. Then,
Tu = (1, 0, −2u),
Tv = (0, 1, 2v)
and thus
N = Tu × Tv = (2u, −2v, 1).
Then the surface area is
∫∫
∫∫
∫∫ √
1 + 4u2 + 4v 2 dA
||N ||dA =
D
D
∫ 2√
∫ 2π
=
1 + 4r2 · rdrdθ
0
0
]2
[
π
1
2 3/2
= (173/2 − 1)
(1 + 4r )
= 2π ·
12
6
0
1dS =
S
5
Problem 8. (10 pts.) Let
→
F =
[
(2x − y − z), (2y − x), (2z − x)
]
→
Assuming that F is conservative, find its potential.
First Solution: Solve the PDE’s:

 ∂x f = 2x − y − z
∂y f = 2y − x

∂z f = 2z − x
Integrating the first equation we get
f (x, y, z) = x2 − (y + z)x + h(y, z)
Putting there g(z) + h(y) instead of more general h(y, z) resulted in huge loss of points. Not
all potentials are of this form and there is no reason to assume that this potential will be.
Computing ∂y f in two ways we get:
∂y f = −x + ∂y h(y, z)
= 2y − x
from the line above
from the system of equations
Thus ∂y h(y, z) = 2y and anti differentiating we get:
h(y, z) = y 2 + g(z) or f (x, y, z) = x2 − (y + z)x + y 2 + g(z)
Lastly computing ∂z f in two ways we get:
∂z f = −x + g ′ (z)
= 2z − x
from the line above
from the system of equations
Thus g(z) = z 2 + C and f (x, y, z) = x2 + y 2 + z 2 − (y + z)x.
Accidentally the final result is of the form that many of you assumed it will be, but assuming
this at the beginning of the problem is not legitimate.
bf Second solution: Let C = Cx ∪ Cy ∪ Cz be the union of three intervals: Cx : from (0, 0, 0)
→
to (a, 0, 0), then Cy to (a, b, 0) and then Cz to (a, b, c). Then the potential (assuming that F
is conservative) is:
(a,b,c)
∫
∫
(2x−y−z)dx+(2y−x)dy+(2z−x)dz =
(2x−y−z)dx+
Cx
(0,0,0)
∫
∫
(2y−x)dy+
Cy
(2z−x)dz
Cz
where we used the fact that e.g. Cx is parallel to x−axis, so that dy = dz = 0. Now on Cx :
y = z = 0; on Cy : x = a and z = 0; on Cz : x = a and y = b, so that:
∫a
f (a, b, c) =
∫b
2y − a dy +
2x dx +
0
∫c
0
2z − a dz = a2 + b2 + c2 − ab − ac
0
6
Problem 9. (10 pts.)
y
3
(4 cos α, 4 sin α )
2
1
C
α
x
−2
−1
−1
1
2
3
Let C be the interval of length 4 that forms the
angle α with the positive x− axis (see the picture
on the left).
a) Use second Pappus theorem to find the area
of the surface S obtained by rotating C about
x−axis.
Your answer should contain the parameter α. For
what angle α is that area maximal?
4
Solution: Pappus theorem states that Area = length of curve × distance traveled by the
centroid.
The length is given and equal to 4 (I have no idea why some of you had to compute it).
By symmetry the centroid is in the middle of the interval, i.e. at the point (2 cos α, 2 sin α ).
Thus the area is
A = 4 · 2π · 2 sin α = 16 sin α
It achieves maximum when sin α = 1, i.e. for α = π/2. (Thus the cone is becomes the disk.)
√
b) For α = π/6, i.e. for the interval C ending at ( 12, 2), parametrize the surface S.
Your parametrization should include the formulas for the x, y and z coordinates as well as
the domains (limits) for the parameters.
√
Solution: The parametrization of the interval ending at ( 12, 2) is:
√
t → ( 12t, 2t)
for
0≤t≤1
To rotate this about the x−axis we need to add the angle θ ∈ [0, 2π ] and send (t, θ) to:
√
( 12t, 2t cos θ, 2t sin θ)
Of course there are many variants of this correct answer. For example one can expand the
domain for t while dividing all coordinates of the parametrization by the same factor. Quite
common on the exam was taking t ∈ [0, 4] and thus sending (t, θ) to:
√
3 t
t
t, cos θ, sin θ)
(
2
2
2
Replacing the locations of sin θ and cos θ results in yet another correct answer.
Important things to look at:
• to parametrize a surface you need exactly two parameters.
7
• in the rotation about x−axis the x coordinate does not change with θ.
• best to solve this problem in two stages: interval first and then the rotation.
Problem 10. (10 pts.) Let T be the region in the first quadrant (x, y > 0) bounded by
the hyperbolas xy = 1, xy = 4 and the lines y = x and y = 25x.
Show that for any continuous function f (t)
∫∫
∫4
f (xy) dA = ln 5 ·
T
f (u) du
1
Hint: use the substitution u = xy, v = y/x.
Solution: With the suggested substitution we have:
∂(u, v) y
y
x = 2 = 2v
=
2
−y/x 1/x
∂(x, y)
x
Thus
∂(x, y)
1
=
∂(u, v)
2v
and
∫4 ∫25
∫∫
f (xy) dA =
T
1
1
25 ∫4
∫4
1
1
f (u)
dv du = ln(v)
f (u) du = ln 5
f (u) du
2v
2
1
1
8
1