Math 392 problems — with Answers
Professor Chavel
To the student: The following problems are lifted from my files. Some errors may have crept
into the text during the copy/paste. Some problems might be repeated. Regrettably, they do
not cover all the topics, especially linear algebra and the curvature of plane curves. My advice:
Copy a batch of problems, do them and then, check the answers and give yourself a grade.
Question. Calculate the line integral
H
C
F · dr, where
F = 3zi − 7xj + 5yk,
and C is the intersection of z = x + 4 with x2 + y 2 = 4.
Answer. One can do it two ways: The first is directly. The curve C is parametrized by
x = 2 cos θ,
y = 2 sin θ,
z = 2 cos θ + 4,
0 ≤ θ ≤ 2π;
so
dr = {−2 sin θi + 2 cos θj − 2 sin θk} dt.
Restricted to the curve C, the vector field F is given by
F = 3(2 cos θ + 4)i − 7 · 2 cos θj + 5 · 2 sin θk.
Now take the dot product of F with dr and integrate from 0 to 2π.
The second way is to use Stokes theorem:
I
ZZ
F · dr =
(∇ × F) · n dA,
C
S
where S is the surface z = x + 4 that is inside the cylinder x2 + y 2 = 4. One has by direct calculation
∇ × F = 5i + 3j − 7k,
which implies
ZZ
n dA = (−i + k) dxdy,
ZZ
(∇ × F) · n dA =
(−5 − 7) dxdy = −12 · 4π = −48π.
x2 +y 2 =4
S
Question. Calculate the line integral
H
C
F · dr, where
F = 3zi − 7xj + 4yk,
and C is the intersection of z = 2y + 7 with x2 + y 2 = 9.
Answer. Again, one has two ways to do it: either directly or with Stokes’ theorem.
To do it directly, one parametrizes the curve C by
r(t) = 3 cos ti + 3 sin tj + (2(3 sin t) + 7)k,
0 ≤ t ≤ 2π,
which implies
dr = −3 sin ti + 3 cos tj + 6 cos tk,
and
F = 3(6 sin t + 7)i − 7(3 cos t)j + 4(3 sin t)k;
so
F · dr = −9 sin t(6 sin t + 7) − 63 cos2 t + 72 cos t sin t,
which one integrates from 0 to 2π, to obtain −117π.
To use Stokes’ theorem one calculates
curl F = 4i + 3j − 7k,
and for the surface z = 2y + 7, with the normal pointing upward we have
n dA = (−2j + k) dx dy.
therefore, by Stokes’ theorem,
I
ZZ
F · dr =
−13 dx dy = −13π(32 ) = −117π.
x2 +y 2 ≤9
C
Question. Calculate the line integral
z = t3 , with 0 ≤ t ≤ 1.
Answer. We have
which implies
Z
Z
xz ds =
C
1
R
C
√
xz ds, where C is the curve: x = 6t, y = 3 2t2 ,
√
y 0 = 6 2t
x0 = 6
Z
p
(6t)(t3 ) x02 + y 02 + z 02 dt =
1
6t4
z 0 = 3t2 ,
Z
p
36 + 72t2 + 9t4 dt = 3
0
0
1
6t4
p
4 + 8t2 + t4 dt
0
The integral is too difficult to bother with. If you come this far, you get full credit.
Question. Evaluate the line integral
R
C
F·dr, where
F = 2xy 3 z 4 i + 3x2 y 2 z 4 j + 4x2 y 3 z 3 k
and C is the curve given by: r(t) = ti + t2 j + t3 k, 0 ≤ t ≤ 2.
Answer. Well,
r(t) = ti + t2 j + t3 k,
r0 (t) = i + 2tj + 3t2 k,
F(r(t)) = 2(t)(t2 )3 (t3 )4 i + 3(t)2 (t2 )2 (t3 )4 j + 4(t)2 (t2 )3 (t3 )3 k
= 2t19 i + 3t18 j + 4t17 k,
0
F(r(t))·r (t) = 2t19 + 6t19 + 12t19 = 20t19 ,
Z
Z 2
F·dr =
20t19 dt = 220 .
C
0
Question. Evaluate the line integral
Z
C
(y + e
√
x
) dx + (2x + cos y 2 ) dy,
where C is the boundary of the region enclosed by the parabolas y = x2 and x = y 2 .
Answer. One can evaluate the integral directly or use Green’s theorem.
To do it directly, first note that
Z
√
e
x
dx + cos y 2 dy = 0
C
where C is any closed curve (why?). So one only has to worry about
Z
y dx + 2x dy.
C
The √
curve C is given in two parts: C1 is the curve y = x2 where x varies from 0 to 1; and C2 is the curve
y = x where x varies from 1 to 0 — note the orientation! Therefore
Z
y dx + 2x dy
C
Z
1
=
Z
0
Z
0
x2 dx + 2x d(x2 ) +
Z
x2 dx + 2x(2x) dx −
0
=
√
x dx + 2x d( x)
1
1
=
Z
√
1
√
0
1
2
1/2
5x − 2x
1
x dx + 2x x−1/2 dx
2
dx
0
5 4
−
3 3
1
= .
3
=
To use Green’s theorem, one has
Z
ZZ
ZZ
√
x
2
(y + e ) dx + (2x + cos y ) dy =
(2 − 1) dx dy =
C
D
dx dy,
D
where D is the region bounded bounded by the curves. To calculate the double integral one uses the
iterated integral, namely,
ZZ
Z (Z 1/2 )
1
dx dy
x
=
D
dy
Z
x2
0
1
=
{x1/2 − x2 } dx
0
=
1
.
3
Question. Calculate the arclength of the path
r(t) = i + t2 j + t3 k,
0 ≤ t ≤ 1.
Answer. One calculates
r(t)
= i + t2 j + t3 k
r0 (t)
=
|r0 (t)|
2t j + 3t2 k
p
=
4t2 + 9t4
p
= t 4 + 9t2
Z
length
1
=
Z
|r0 (t)| dt
0
1
=
t
p
4 + 9t2 dt
0
Question. Calculate the line integral
=
¯1
(4 + 9t2 )3/2 ¯¯
¯
27
0
=
(13)3/2 − 8
.
27
Z
xeyz ds,
over the line segment from (0, 0, 0) to (1, 2, 3).
Answer. First parameterize the path by
x = t,
Then
which implies
y = 2t,
2
xeyz = te6t ,
z = 3t,
ds =
Z
xeyz ds =
√
Z
0 ≤ t ≤ 1.
p
12 + 22 + 32 dt =
1
14
√
14 dt,
√
2
te6t dt =
0
14 6
{e − 1}.
12
Question. Calculate the line integral
Z
z dx + x dy + y dz,
over the path: x = t2 , y = t3 , z = t2 , where 0 ≤ t ≤ 1.
Answer. Calculate
Z
Z
z dx + x dy + y dz
1
=
{(t2 )(2t) + (t2 )(3t2 ) + (t3 )(2t)} dt
0
Z
=
1
{2t3 + 5t4 } dt
0
=
1
3
{ + 1} = .
2
2
Question. Calculate the line integral
Z
F· dr,
F = x sin y i + y j,
over the path: y = x2 , from (−1, 1) to (2, 4).
Answer. So −1 ≤ x ≤ 2, and
r(x)
= xi + x2 j
0
r (x) = i + 2xj
F = x sin x2 i + x2 j
F· dr
= {x sin x2 + 2x3 } dx,
which implies
Z
Z
2
F· dr =
{x sin x2 + 2x3 } dx
−1
¯2
cos x2
x4 ¯¯
−
+ ¯
2
2 −1
1
{cos 1 − cos 4 + 16 − 1}.
2
=
=
Question. Calculate the line integral
I
F = x2 i + xy j,
F· dr,
over the circle: x2 + y 2 = 4.
Answer. The circle is parameterized by
r(t) = 2(cos t)i + 2(sin t)j,
which implies
F· dr
= {4(cos2 t)i + 4(cos t sin t)j}·{−2(sin t)i + 2(cos t)j}
= 8{− sin t cos2 t + sin t cos2 t} dt = 0.
So the integral is equal to 0.
Question. Calculate the line integral
Z
F = y 2 cos z i + 2xy cos z j − xy 2 sin z k,
F· dr,
over the path in R3 : x = t2 , y = t + 1, z = 2t − 1, where 0 ≤ t ≤ 1.
Answer. The secret here is to realize that F is the gradient of a function, namely
F = grad xy 2 cos z.
Therefore the integral is evaluate by evaluating the function φ = xy 2 cos z at the two endpoints and
subtracting:
Z
¯(1,2,1)
F· dr = xy 2 cos z ¯(0,1,−1) = 4 cos 1.
Question. Calculate the surface integral
that lies inside the cylinder x2 + y 2 = 1.
RR
S
yz dA where S is the part of the plane z = y + 3
Answer. The area element of the surface z = y + 3 is given by
q
√
dA =
zx 2 + zy 2 + 1 dxdy = 2 dxdy,
which implies
ZZ
yz dA =
S
√ ZZ
2
(y 2 + 3y) dxdy.
x2 +y 2 ≤1
Because of symmetry relative to the x–axis, the contribution of 3y to the integral is 0. For the contribution
of y 2 one can use polar coordinates:
Z 1
√ ZZ
√ Z 2π
√
2
y 2 dxdy = 2
sin2 θ dθ
r2 r dr = 2π/4.
x2 +y 2 ≤1
0
RR
Question. Calculate the surface integral
S
0
F · n dA, where
F = xi + yj + z 2 k,
and S is the part of the paraboloid of revolution z = x2 + y 2 beneath the plane z = 2 with
downward orientation.
Answer. In our case the surface is given in the form z = ϕ(x, y), where
ϕ = x2 + y 2 ,
and the normal vector is pointing downward, so
½
¾
∂ϕ
∂ϕ
n dA =
i+
j − k dx dy
∂x
∂y
= {2xi + 2yj − k} dx dy.
Therefore
ZZ
ZZ
{2x2 + 2y 2 − (x2 + y 2 )2 } dx dy.
F · n dA =
x2 +y 2 ≤2
S
To evaluate the integral use polar coordinates; so
ZZ
√
Z
F · n dA =
2
2π
S
Z
=
0
2π
√
2
(2r2 − r4 )r dr
(2r3 − r5 ) dr
0
½
=
2π
=
4π
.
3
Question. Calculate the surface integral
RR
S
2(21/2 )4
(21/2 )6
−
4
6
¾
F · n dA, where
F = xi + yj + z 2 k
and S is the part of the cone z =
Answer. Do it directly. z =
p
p
x2 + y 2 beneath the plane z = 1 with downward orientation.
x2 + y 2 implies
x
n dA = p
x2
and
+
y2
y
i+ p
x2
+ y2
F = xi + yj + {x2 + y 2 }k,
j − k,
which implies
ZZ
(
ZZ
F · n dA =
x2 +y 2 =1
S
ZZ
=
x2 + y 2
p
− (x2 + y 2 )
x2 + y 2
)
dxdy
n
o
{x2 + y 2 }1/2 − (x2 + y 2 ) dxdy
x2 +y 2 =1
Z 1
=
=
{r − r2 }r dr
2π
π
.
6
0
Question. Calculate the surface integral
RR
S
F · n dA, where
F = x3 i + 2xz 2 j + 3y 2 zk
and S is surface of the solid bounded by the paraboloid z = 4 − x2 − y 2 and the xy–plane.
Answer. Call the solid D. Its boundary, ∂D, is a surface which has no boundary, and is composed of
two pieces. The first is the paraboloid below z = 4, and the second is the circle x2 + y 2 ≤ 4 in the plane
z = 0. So the surface integral would have to be done in two parts. Therefore, it seems simplest to use
the divergence theorem
ZZ
ZZZ
F · n dA =
div F dV.
∂D
First
D
div F = 3x2 + 3y 2 = 3(x2 + y 2 ),
which suggests we should do the triple integral with cylindrical coordinates, that is, the coordinates in
the xy–plane should be polar coordinates. Then
(Z (Z
)
)
2
ZZZ
Z
2π
div F dV
=
D
0
Z
=
2
4−r
r2 dz
3
0
2
6π
r dr
dθ
0
{4 − r2 }r3 dr
0
Z
=
6π
2
{4r3 − r5 } dr = 32π.
0
Question. For a domain D in three-space R3 , show that the volume of D, V (D), is given by
the formula
ZZ
1
V (D) =
r · n dA,
3 ∂D
where ∂D denotes the boundary of D, dA the infinitesmal surface area element on ∂D, n the
exterior unit normal vector field along ∂D, and r the position vector considered as a vector field
r(x, y, z) = xi + yj + zk.
Answer. Use the divergence theorem.
ZZ
ZZZ
r · n dA =
∂D
ZZZ
div r dV =
D
3 dV = 3V (D),
D
which implies the formula.
RR
Question. Calculate the surface integral
F·n dA, where
S
F = xi + yj + z 4 k
and S is the part of the cone z =
p
x2 + y 2 beneath the plane z = 1 with downward orientation.
Answer. With downward orientation,
and the surface given as z = φ(x, y), we have, for the normal-area
p
element of the surface z = x2 + y 2 ,
n dA =
{φx i + φy j − k} dxdy
(
)
x
y
p
i+ p
j − k dxdy,
x2 + y 2
x2 + y 2
=
F(φ(x, y))
xi + yj + {x2 + y 2 }2 k,
)
(
y2
x2
2
2 2
p
+p
− {x + y }
dxdy,
x2 + y 2
x2 + y 2
p
x2 + y 2 − {x2 + y 2 }2 ,
=
F(φ(x, y))·n dA =
=
which implies, using polar coordinates,
ZZ
F·n dA
Z
1
= 2π
S
{r − r4 }r dr
0
= 2π{1/3 − 1/6} = π/3.
RR
Question. Calculate the surface integral
F · n dA, where
S
F = x3 i − 4x2 z 2 j + 3y 2 zk,
and S is boundary surface of the solid bounded by the upper hemisphere z =
the xy–plane.
p
4 − x2 − y 2 and
Answer. Here one definitely works with the divergence theorem. The divergence of F is given by
div F = 3x2 + 3y 2 .
So to calculate the triple integral of div F in the solid upper hemisphere, we introduce usual polar
coordinates in the xy–plane. Then
div F = 3r2 ,
and
ZZZ
√ 2 2
4−x −y ,
(Z √ 2
{0≤z≤
Z
2
4−r
div F dx dy dz
x2 +y 2 ≤4}
)
2
3r dz
= 2π
Z
= 6π
r dr
0
0
2
r3
p
4 − r2 dr.
0
One calculates this integral by picking the substitution
ρ2 = 4 − r 2
⇒
ρ dρ = −r dr,
from which one has
Z
2
6π
r
0
3
p
Z
4−
r2
dr = 6π
2
½
2
4
(4ρ − ρ ) dρ = 6π
0
32 32
−
3
5
¾
=
128π
.
5