Solutions to Vector Calculus Practice Problems
1. Let R be the region in R2 determined by the inequalities x2 + y 2 ≤ 4
and y 2 ≤ x2 . Evaluate the following integral.
ZZ
sin(x2 + y 2 ) dA
R
Answer: The region looks like
y
y=x
x
y = —x
We use polar coordinates:
Z
π/4
Z
Z
2
2
5π/4
Z
2
r sin(r ) dr dθ +
−π/4
0
3π/4
r sin(r2 ) dr dθ
0
=
¶
¶
Z 5π/4 µ
π/4 µ
1
1
1
1
− cos(4) +
dθ +
dθ
− cos(4) +
2
2
2
2
−π/4
3π/4
=
π
π
− cos(4) +
2
2
Z
1
ZZZ
2. Evaluate
z dV , where R is the following solid region:
R
z
(0, 0, 1)
(1, 0, 1)
(0, 1, 1)
(0, 1, 0)
(2, 0, 0)
y
(1, 1, 0)
x
Answer: If we integrate with respect to x first, we will not need to
break the region into more than one piece (that is, we will only need
to use one integral). The plane containing the front parallelogram has
equation x + y + z = 2. (You should be able to find this by the “guessand-check” method.) So, the bounds for x are 0 ≤ x ≤ 2 − y − z. We
are integrating over a square in the yz-plane, so the y and z bounds
are 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1. Thus:
ZZZ
Z 1 Z 1 Z 2−y−z
Z 1Z 1
¡
¢
z dV =
z dx dy dz =
2z − yz − z 2 dy dz
R
Z
0
0
1µ
=
0
0
2 1
− y
3 2
0
¶
dz =
0
5
12
3. Let R be the region in R3 determined by the Z
inequalities
r ≤ 1 and
ZZ
p
0 ≤ z ≤ 4 − r2 , where r = x2 + y 2 . Evaluate
r dV .
R
Answer: R is the region under a paraboloid, inside a cylinder, and
above the xy-plane. We will use cylindrical coordinates.
ZZZ
Z 2π Z 1 Z 4−r2
Z 2π Z 1
¡
¢
2
r dV =
r dz dr dθ =
r2 4 − r2 dr dθ
R
Z
0
2π
=
0
·
0
0
4 3 1 4
r − r
3
5
2
¸1
0
Z
2π
=
0
0
0
17
34π
dθ =
15
15
4. Let R be the region in R3 defined by 4 ≤ x2 + y 2 + z 2 ≤ 9 and z ≥ 0.
Evaluate the following integral:
ZZZ
¡ 2
¢
x + y 2 dV
R
Answer: R is the region inside a sphere of radius 3, outside a sphere
of radius 2, and above the xy-plane. We will use spherical coordinates,
because the region is very easy to describe in spherical coordinates:
π
2 ≤ ρ ≤ 3, 0 ≤ φ ≤ , 0 ≤ θ ≤ 2π. Since r = ρ sin φ, we are
2
integrating x2 + y 2 = r2 = ρ2 sin2 φ. Thus:
ZZZ
¡
2
x +y
2
¢
Z
dV
Z
2π
π/2
Z
3
=
R
Z
0
2π Z
0
π/2 Z
0
Z
2π
0
3
211
=
5
0
Z
ρ4 sin3 φ dρ dφ dθ
2
π/2
=
0
¢
ρ2 sin2 φ ρ2 sin φ dρ dφ dθ
2
=
Z
¡
2π
0
211 3
sin φ dρ dφ dθ
5
Z
π/2
¡
¢
1 − cos2 φ sin φ dφ dθ
0
We use the substitution u = cos φ, du = − sin φ dφ:
ZZZ
¡
R
2
x +y
2
¢
dV
211
=
5
Z
2π
0
Z
0
(1 − u2 ) du dθ =
1
844π
15
5. Let C be the curve x = 1 − y 2 from (0, −1) to (0, 1). Evaluate
Z
y 3 dx + x2 dy
C
Answer: We parameterize the curve using t = y:
x = 1 − t2
y = t
3
−1≤t≤1
Then:
dx = −2t dt
dy = dt
Thus:
Z
Z
3
2
y dx + x dy =
C
·
=
1
³
¡
3
t (−2t) + 1 − t
−1
2
1
− t5 − t3 + t
5
3
6. Let C be the curve x =
¢
2 2
Z
´
1
dt =
¡
¢
−t4 − 2t2 + 1 dt
−1
¸1
=
1
4
15
√
t, y = 1 + t3 for 0 ≤ t ≤ 1. Evaluate
Z
x3 y 4 dx + x4 y 3 dy
C
Answer: The rotation of this
¯
¯ ∂
∂
¯
¯ ∂x
∂y
¯
¯ 3 4 4 3
¯ xy xy
vector field equals 0:
¯
¯
¯
¯
¯ = 4x3 y 3 − 4x3 y 3 = 0
¯
¯
This means that the vector field is conservative, so there is some function f with ∇f = x3 y 4 i+x4 y 3 j. It is fairly easy to see that f = 14 x4 y 4 + C
works. The endpoints of the curve C are (0, 1) (when t = 0) and (1, 2)
(when t = 1). Thus:
Z
·
1
x y dx + x y dy = x4 y 4
4
C
3 4
¸(1,2)
4 3
=4−0= 4
(0,1)
Note: This problem can also be done as a standard vector line integral,
but the calculations are somewhat tedious.
4
7. Let C be the circle x2 + y 2 = 4, oriented counterclockwise. Use Green’s
Theorem to evaluate the following integral
I
¡
¢
cos(x2 ) − y 3 dx + x3 dy
C
Answer: First, we compute the rotation of the vector field:
¯
¯
¯
∂
∂ ¯
¯
¯
¯
¯
∂x
∂y
¯
¯ = 3x2 + 3y 2
¯
¯
¯ cos(x2 ) − y 3 x3 ¯
Then, by Green’s Theorem:
I
ZZ
¡
¢
¡ 2
¢
2
3
3
cos(x ) − y dx + x dy =
3x + 3y 2 dA
C
R
where R is the region inside the circle. Using polar coordinates, we
have:
I
Z 2π Z 2
¡
¢
2
3
3
cos(x ) − y dx + x dy =
3r3 dr dθ = 24π
C
0
5
0
8. The following picture shows the parametric curve (x, y) = (t − t3 , t2 ):
y
2
(0, 1)
x
—1
1
Use Green’s Theorem to find the area of the shaded region.
Answer: First, we need to find the bounds for t. In particular, we want
to know when the curve passes through the the point (0, 1). Setting
t − t3 = 0 and t2 = 1 and solving, we get t = −1 and t = 1. Thus, our
bounds are −1 ≤ t ≤ 1.
By plotting a few points of the parametric curve, we can see that the
curve is oriented counterclockwise, which agrees with the orientation
given in Green’s Theorem.
We need to find a vector field F such that rot(F) = 1. The vector field
F = −y i will work (as would x j or lots of other possibilities).
Then, by Green’s Theorem:
ZZ
I
dA =
R
−y dx
C
where R is the shaded region and C is the curve. We have:
x = t − t3
y = t2
for −1 ≤ t ≤ 1. Thus:
ZZ
Z
dA =
R
dx = (1 − 3t2 ) dt
dy = 2t dt
1
¡
¢
8
−t2 1 − 3t2 dt =
15
−1
6
9. Let S be the surface given by the following parametric equations
x = 4t2 + u
y = cos t
z = sin u
Find the equation for the plane tangent to the surface at the point
(π 2 , 0, 0).
Answer: Since we are given the surface by parametric equations, we
can find the normal vector by computing Tt × Tu , and then we can use
the normal vector to find the equation for the plane.
Tt = (8t, − sin t, 0)
Tu = (1, 0, cos u)
¯
¯
¯ i
¯
j
k
¯
¯
0 ¯¯ = − sin t cos u i − 8t cos u j + sin t k
Tt × Tu = ¯¯ 8t − sin t
¯ 1
0
cos u ¯
We can determine the values of t and u at the point (2, 1, 0). At the
point (π 2 , 0, 0), 4t2 + u = π 2 , cos t = 0, and sin u = 0. Solving, we get
t = π/2 and u = 0. Thus, the normal vector at this point is:
³π ´
Tt × Tu
, 0 = −i − 4π j + k
2
Thus, the equation for the plane is of the form −x − 4πy + z = D.
We can solve for D by plugging in the point (π 2 , 0, 0). We get that
−π 2 = D. Thus, the equation for the plane is
−x − 4πy + z = −π 2
10. Let S be the surface z = x2 + y 2 , 0 ≤ z ≤ 4. Evaluate the following
integral:
ZZ
x
√ dA
z
S
7
Answer: The surface is a paraboloid. It will be easiest to work this
problem if we use the parameters t = r and u = θ. Then, we can
parameterize the surface as follows:
x = t cos u
y = t sin u
z = t2
0≤t≤2
0 ≤ u ≤ 2π
The tangent vectors to the surface are
Tt = (cos u, sin u, 2t)
Tu = (−t sin u, t cos u, 0)
Then, the normal vector to the
¯
¯
i
j
¯
¯
sin u
Tt × Tu = ¯ cos u
¯ −t sin u t cos u
Thus:
surface is
¯
k ¯¯
2t ¯¯ = −2t2 cos u i − 2t2 sin u j + t k
0 ¯
p
dA = kTt × Tu k dt du = 4t4 cos2 +4t4 sin2 u + t2 dt du
√
√
4t4 + t2 dt du = t 4t2 + 1 dt du
=
We can now compute the surface integral:
ZZ
Z 2π Z 2
x
t cos u √ 2
√ dA =
√ t 4t + 1 dt du
z
t2
S
0
0
Z 2π Z 2
√
t cos u 4t2 + 1 dt du
=
0
0
2
We use the substitution v = 4t + 1, dv = 8t dt:
Z Z
ZZ
√
x
1 2π 17
√ dA =
cos u v dv du = 0
8 0
z
S
1
Note: It is also possible to see that the integral is 0 without doing
any computations, since
√ the region is symmetric about the yz-plane,
as is the function x/ z. This is similar to how the integral of an odd
function from −a to a is 0.
8
11. Let S be the surface z =
integral:
ZZ
p
x2 + y 2 , 1 ≤ z ≤ 2. Evaluate the following
¡
¢
x i + y j + z 2 k · dA
S
Answer: We will parameterize this surface with parameters t = r and
u = θ. Then, the parametric equations are
x = t cos u
y = t sin u
z = t
1≤t≤2
0 ≤ u ≤ 2π
The tangent vectors to the surface are
Tt = (cos u, sin u, 1)
Tu = (−t sin u, t cos u, 0)
The, the normal vector to the surface is
¯
¯
¯
i
j
k ¯¯
¯
sin u 1 ¯¯ = −t cos u i − t sin u j + t k
Tt × Tu = ¯¯ cos u
¯ −t sin u t cos u 0 ¯
Now, we compute F · (Tt × Tu ):
¡
¢
F · (Tt × Tu ) = x, y, z 2 · (−t cos u, −t sin u, t)
¡
¢
= t cos u, t sin u, t2 · (−t cos u, −t sin u, t)
= −t2 cos2 u − t2 sin2 u + t3 = −t2 + t3
Now, we can compute the surface integral:
ZZ
Z 2π Z 2
¡
¢
2
F · (Tt × Tu ) dt du
x i + y j + z k · dA =
S
Z
0
2π Z
1
2
=
0
1
¡ 2
¢
17π
−t + t3 dt du =
6
Note: This problem should have specified an orientation. The above
answer corresponds to unit normals oriented pointing inwards towards
9
the z-axis (you can see this from the vector Tt × Tu ). The answer
17π
would be −
if we oriented the surface with unit normals pointing
6
outwards.
12. Let S be the union of the cylinder x2 + y 2 = 4 for −3 ≤ z ≤ 3 and the
2
2
2
hemisphere
Z Z x + y + (z − 3) = 4 for z ≥ 3. Use Stokes’s Theorem to
¡
¢
evaluate
2yz j − z 2 k · dA.
S
Answer: In order to use Stokes’s Theorem, we need to compute the
anti-curl of 2yz j − z 2 k. The vector field F = yz 2 i will work. Then:
ZZ
I
¡
¢
2
2yz j − z k · dA =
yz 2 i · ds
S
C
where C is the boundary of the surface. The boundary of the surface
is the circle x2 + y 2 = 4 in the plane z = −3. We can parameterize this
curve:
x = 2 cos t
y = 2 sin t
0 ≤ t ≤ 2π
z = −3
Then:
dx = −2 sin t dt
dy = 2 cos t dt
dz = 0
Thus:
ZZ
I
Z
¡
¢
2
2
2yz j − z k · dA =
yz dx =
S
Z
C
Z
µ
2π
2
−36 sin t dt =
0
=
18 sin t(−2 sin t) dt
0
2π
=
2π
−36
0
1 − cos(2t)
2
−36π
Note: The region should have been oriented. The given answer is
correct if the region is oriented with outward pointing normals. If the
normals pointed inwards, the answer would be 36π.
10
¶
dt
13. Let C be the the rectangle in R3 with vertices (0, 0, 0), (1, 0, 0), (1, 1, 1),
and (0, 1, 1), oriented in the given order. Use Stokes’s Theorem to
evaluate the following integral:
I
sin(x2 ) dx + xy 2 dy + xz 2 dz
C
Answer: Using Stokes’s Theorem, we can change the line integral into
a surface integral over a surface whose boundary is the given curve. One
such surface is the interior of the given rectangle. The plane containing
the four points has equation y − z = 0, so we can parameterize the
surface using the parameters t = x and u = y. Since the rectangle
lies over the unit square in the xy-plane, the bounds for x and y are
0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
x = t
y = u
z = u
0≤t≤1
0≤u≤1
Then, according to Stokes’s Theorem:
I
ZZ
2
2
2
sin(x ) dx + xy dy + xz dz =
(∇ × F) · dA
C
S
where S is the surface parameterized above and where F = sin(x2 ) i +
xy 2 j + xz 2 k. We can compute ∇ × F:
¯
¯
¯
i
j
k ¯¯
¯
¯
∂
∂ ¯¯
∂
¯
∇×F=¯
¯ = −z 2 j + y 2 k
¯ ∂x
∂y
∂z ¯
¯
¯
¯ sin(x2 ) xy 2 xz 2 ¯
To compute the surface integral, we need to find Tt × Tu :
Tt = (1, 0, 0)
Tu = (0, 1, 1)
So, Tt ×Tu = (0, 1, −1). This vector points downwards, which disagrees
with the orientation of the rectangle (which is oriented counterclockwise
11
about upwards pointing vectors), so we need to negate the resulting
integral. Thus:
ZZ
Z 1Z 1
¡ 2
¢
(∇ × F) · dA = −
−z j + y 2 k · (0, 1, −1) dt du
S
Z
0
1
Z
1
Z
0
1
¡ 2
¢
−z − y 2 dt du
1
¡
¢
2
−2u2 dt du =
3
= −
Z
0
0
= −
0
0
14. Let S be the boundary of the region x2 + y 2 ≤ 4, 0 ≤ z ≤ 3, oriented
with unit normals pointing outwards. Consider the vector field
¡
¢
¡
¢
F = x3 + cos(y 2 ) i + yz j + 3y 2 z + cos(xy) k
Use the divergence theorem to evaluate the following integral:
ZZ
F · dA
S
Answer: The region is a cylinder. Note that the region is oriented appropriately to apply the divergence theorem. Let us denote the region
by R. Using the divergence theorem, we have
ZZ
ZZZ
F · dA =
(∇ · F) dV
S
R
We can compute the ∇ · F:
∇ · F = 3x2 + z + 3y 2
We can set up the integral in cylindrical coordinates:
ZZ
Z 2π Z 2 Z 3
¡ 2
¢
3r + z r dz dr dθ = 90π
F · dA =
S
0
0
0
12
15. Let S be the surface in R3 defined by the parametric equations
r = 2 + cos t + cos u
θ = t
0 ≤ t ≤ 2π
0 ≤ u ≤ 2π
z = sin u
Use the divergence theorem to find the volume of the region inside of S.
Answer: In order to use the divergence theorem, we need to compute
the anti-divergence of the constant function 1. Some simple vector
fields that work are x i, y j, and z k. We will use z k, as it will make
the computations easier.
We need to parameterize the surface in terms of x, y, and z. We use
the same parameters t and u as above, and the fact that x = r cos θ
and y = r sin θ. Then:
x = (2 + cos t + cos u) cos t
y = (2 + cos t + cos u) sin t
z = sin u
0 ≤ t ≤ 2π
0 ≤ u ≤ 2π
Note that the equation for z is simpler than the equation for x and y;
this is why we are using the vector field z k.
The tangent vectors to the surface are
¡
¢
Tt = −2 sin t − cos u sin t, 2 cos t + cos2 t + cos u cos t − sin2 t, 0
Tu = (− sin u cos t, − sin u sin t, cos u)
The normal vector to the surface is
¯
¯
i
j
¯
2
¯
Tt × Tu = ¯ −2 sin t − cos u sin t 2 cos t + cos t + cos u cos t − sin2 t
¯
− sin u cos t
− sin u sin t
k
0
cos u
¯
¯
¯
¯
¯
¯
Since we will be computing (Tt × Tu ) · z k, we only need to know the k
component of Tt × Tu .
2 sin2 t sin u+sin2 t sin u cos u+2 cos2 t sin u+cos3 t sin u+cos u cos2 t sin u−sin3 u cos t
13
We can simplify this some using cos2 t + sin2 t = 1. We get:
2 sin u + cos3 t sin u + cos u sin u − sin3 u cos t
Thus:
¡
¢
(Tt × Tu ) · z k = sin u 2 sin u + cos3 t sin u + cos u sin u − sin3 u cos t
= 2 sin2 u + cos3 u sin u + cos u sin2 u − sin4 u cos t
Thus:
ZZZ
ZZ
Volume =
Z
dV =
R
Z
2π
2π
=
0
¡
Z
2π
zk · dA =
S
Z
2π
z k · (Tt × Tu ) dt du
0
0
¢
2 sin2 u + cos3 u sin u + cos u sin2 u − sin4 u cos t dt du
0
Z
2π
= 2π
¡
¢
2 sin2 u + cos3 u sin u + cos u sin2 u du
0
1
To integrate the first term, we use the trig identity sin2 u = (1 − cos(2u)).
2
To integrate the second term, we use the substitution v = cos u, dv =
− sin u du. To integrate the third term, we use the substitution w =
sin u, dw = cos u du.
Z 2π
Z 1
Z 0
3
Volume = 2π
(1 − cos(2u)) du + 2π
−v dv + 2π
w2 dw
0
1
0
= 4π 2 + 0 + 0 = 4π 2
Since the answer is positive, we know that the surface was oriented
appropriately.
14