First practice midterm – Solutions
1. Let C be the curve defined as follows:
� 2
t
0�t�1
x(t) =
cos(2πt) 1 � t � 3
�
(t − 2)et + 3 0 � t � 2
y(t) =
t+1
2�t�3
�
10
z(t) =
0�t�3
1 + t2
Find
�
(2x + y cos(xy)) dx + (x cos(xy)) dy + 2 dz
C
Solution: This is a complicated curve that starts at
(0, 1, 10) and goes to (1, 4, 1). Our best hope
is that
�
this is a path independent integral, i.e., C ∇f · r ds.
If this were true then
∂f
= 2x + y cos(xy)
∂x
∂f
= x cos(xy)
∂y
∂f
=2
∂z
2. Let
� G be the surface corresponding to the cone
z = x2 + y2 above the region in the plane satisfying
1 � x2 + y2 � 9. Given that the density at a point
is given by δ(x, y, z) = z, find the total mass of the
surface.
Solution: To compute the mass we need to compute
a surface integral, i.e.,
��
δ(x, y, z)d(SA).
G
Let R be the region in the plane 1 � x2 + y2 � 9. Also
note that
�
d(SA) = (gx )2 + (gy )2 + 1 dA
��
�2 �
�2
x
y
�
�
=
+
+ 1 dA
x2 + y2
x2 + y2
√
= 2 dA.
Therefore we have
�� √
�� �
√
z 2 dA =
x2 + y2 2 dA
G
Integrating the first term with respect to x we can
conclude
f = x2 + sin(xy) + C(y, z).
Taking the derivative of this with respect to y and
comparing to the second term we would want
G
Given the nature of the region we are integrating
over, we naturally should work in polar coordinates.
Therefore we have
�� �
√
√ � 2π � 3 2
2
2
x + y 2 dA = 2
r dr dθ
G
0
∂f
∂C
= x cos(xy) +
= x cos(xy)
∂y
∂y
from which we can conclude that
C(y, z) = D(z), i.e.,
∂C
∂y
= 0 so that
f = x2 + sin(xy) + D(z).
Taking the derivative of this with respect to z and
comparing to the third term we would want
∂f
= D � (z) = 2
∂z
from which we can conclude that D(z) = 2z, i.e.,
f = x2 + sin(xy) + 2z.
Now the integral is easy, namely we evaluate it at
the end point and subtract what we get when we
evaluate at the starting point, i.e.,
�
(2x + y cos(xy)) dx + (x cos(xy)) dy + 2 dz
C
�
� �
= f(1, 4, 1)−f(0, 1, 10) = 1+sin(4)+2 − 0+sin(0)+20)
= sin(4) − 17.
1
�
√ � 2π 1 3 �r=3
= 2
r ��
dθ
0 3
r=1
√ �
26 2 2π
=
dθ
3
0
√
52 2π
=
.
3
3. Let S be the region in the plane above the x − axis,
inside the circle of radius 4 centered at the origin and
outside the circle of radius 1 centered at (0, 2). Find
�
�
�
�
�
xy + cos(x2 ) dx + x − arctan(y2 ) dy
∂S
where the boundaries are oriented so that when traveling along the curve the region is to the left.
Solution: Applying Green’s Theorem we have
�
�
�
�
�
xy + cos(x2 ) dx + x − arctan(y2 ) dy
∂S
��
��
��
=
(1 − x) dA +
dA −
x dA
S
S
S
The second integral will be zero by symmetry, the
first integral is the area, which is the area of half a
circle of radius 4 (so 8π) minus the area of a circle of
radius 1 (so π). So we can conclude
�
�
�
�
�
xy + cos(x2 ) dx + x − arctan(y2 ) dy = 7π.
∂S
4. Let S be the solid cube with one corner at (0, 0, 0)
and the opposite corner at (1, 1, 1). Given that
�
�
F = x3 y, sin(z2 ) − 2yz, z2 + 137x6
find
��
∂S
S
=
=
�1 �1 �1
0
�1 �1
0
=
3x2 y dz dy dx
0 0 0
�1 �1
0
=
�1
0
�1
�z=1
�
3x2 yz��
dy dx
z=0
3x2 y dy dx
0
�y=1
3 2 2 ��
x y �
dx
2
y=0
3 2
x dx
0 2
�x=1
1 �
= x3 ��
2
=
�
�
∇ × xy − z + x7 , 137y − x2 + zy, y − xz
�
�
i
�
�
∂
= ��
∂x
�
� xy − z + x7
F · n d(SA).
Solution: Since this is a solid we can either work
through each of the six faces and compute the surface
integral on each, or we can use divergence theorem.
We will opt for the latter. So we can conclude
��
���
F · n d(SA) =
(3x2 y − 2z + 2z) dV
∂S
have a surface with boundary. As with most of our
Stokes’s Theorem problems we can simplify matters
immensely by first noting that instead of working
with the cylinder we can opt to work with a surface
with the same boundary. In our case it will be the region in the xy-plane (i.e., z = 0) with −2 � x � 2
and 0 � y � 3, we will denote this region by T .
Now as for orientation, since we traverse the curve
clockwise when viewed from above, that means our
normal vector will point down, i.e., for T we have
n = �0, 0, −1�. Finally we note that
x=0
1
= .
2
5. Take the solid cylinder x2 + z2 � 4 with 0 � y � 3.
Let G be the part of the boundary of the cylinder with
z � 0. Find
�
� �
7
2
xy − z + x , 137y − x + zy, y − xz · T ds
∂G
where ∂G is oriented clockwise when viewed from
above.
Solution: “Solid” is misleading in this problem,
this really a Stokes’s Theorem problem because we
j
∂
∂y
137y − x2 + zy
k
∂
∂z
y − xz
�
�
�
�
�
�
�
�
= �1 − y, −1 − z, −2x − x� = �1 − y, −1 − z, −3x�
Putting everything together we have
�
�
�
xy − z + x , 137y − x + zy, y − xz · T ds
�
�
�
7
2
=
xy − z + x , 137y − x + zy, y − xz · T ds
∂T
��
�
� �
�
=
1 − y, −1 − z, −3x · 0, 0, −1 dA
T
��
=
3x dA = 0
∂G
7
2
T
The last step following by symmetry of the region.
Second practice midterm – Solutions
1. Let C = (sin t, cos t) for 0 � t � π. Find
�
x dx + x2 y dy.
C
Solution: This is not conservative. Ugh. We put
everything in terms of t and we get
�
�π
2
x dx + x y dy =
sin t cos t + sin2 t cos t(− sin t) dt
C
0
�t=π
�
1
1
2
4 �
= ( sin t − sin t)�
2
4
t=0
= 0.
2. Consider a wire that is in the shape given by the
curve
�
�
(x, y) = t3 − 3t + 2, 12 − 3t2
for −2 � t � 2. Given that the the wire is homogeneous, i.e., δ(x, y) = 1 find the moment of the wire
with respect to the y-axis.
3. Let C be the curve oriented counterclockwise consisting of the semicircle of radius 3 with center at
the origin above the x-axis, the semicircle of radius
2 with center at the origin below the x-axis and the
line segments from (2, 0) to (3, 0) and from (−2, 0) to
(−3, 0). Find
�
C
�
�
cos(x) + 2xy, x − y5 · T ds
Solution: Applying the variation of Green’s Theorem we have
�
��
�
�
cos(x) + 2xy, x − y5 · T ds =
(1 − 2x) dA
S
C
where S is the region enclosed in the curve. The first
part of the integral (i.e., the 1) will find the area, the
second part of the integral (i.e., the −2x) will be zero
by symmetry. Since the area is half of a circle of radius 3 (so 92 π) and half of a circle of radius 2 (so 2π)
we have the total area is 13
2 π and we can conclude
�
Solution: We need to do a line integral, in particular we need to compute
�
xδ(x, y) ds.
C
�
�
13
cos(x) + 2xy, x − y5 · T ds =
π
2
C
To do this we first note that
�
�
ds = (x � )2 + (y � )2 dt = (3t2 − 3)2 + (−6t)2 dt
�
�
= 9t4 − 18t2 + 9 + 36t2 dt = 9t4 + 18t2 + 9 dt
�
�
�2
=
3(t2 + 1) dt = 3(t2 + 1) dt.
4. Let S be the solid cone consisting
� of the points
below the plane z = 4 and above z = x2 + y2 . Find
��
�
�
F = xy2 + ey+cos y , x2 y + sin z, z2 + cos x
−2
=
�2
3t5 − 6t3 + 6t2 − 9t + 6 dt
−2
=
�2
F · n d(SA),
where
So we have
�
�2
xδ(x, y) ds =
(t3 − 3t + 2)3(t2 + 1) dt
C
∂S
2
6t + 6 dt
−2
�
�
��t=2
= 2t2 + 6t ��
t=−2
= 20 − (−20) = 40.
(In going from the second to the third line we observe
that any odd power would be zero when integrating
from −2 to 2.)
Solution: This is a solid and so we use Divergence
Theorem. In particular we have
��
∂S
F · n d(SA)
���
���
=
(∇ · F) dV =
(y2 + x2 + 2z) dV
S
S
Given that this is a cone it is easy to express the
integral in cylindrical coordinates (i.e., we for from
z=
�
x2 + y2 = r to z = 4). Now we have
���
(y2 +x2 +2z) dV =
S
=
=
=
� 2π � 4
0
� 2π
0
� 2π
0
0
�4
� 2π � 4 � 4
0
(r2 +2z)r dz dr dθ
0
r
�z=4
�
(r3 z + z2 r)��
dr dθ
z=r
(3r3 + 16r − r4 ) dr dθ
0
�r=4
3 4
1 5 ��
2
( r + 8r − r )�
dθ
4
5
r=0
�
576 2π
1152π
=
dθ =
.
5 0
5
5. Let G be the surface z = x2 − y2 intersecting the
cylinder x2 + y2 � 1. Find
�
�
�
x, x + y3 , x2 + y2 − z · T ds
∂G
where ∂G is oriented counterclockwise when viewed
from above.
Solution: By Stoke’s Theorem this is equal to
��
�
�
(∇ × x, x + y3 , x2 + y2 − z ) · n d(SA).
G
We have
�
�
∇ × x, x + y3 , x2 + y2 − z
�
� i
j
k
�
� ∂
∂
∂
= ��
∂x
∂y
∂z
�
� x x + y3 x2 + y2 − z
�
�
�
�
� = �2y, −2x, 1�
�
�
�
Also recall that if z = g(x, y) over a region R in
the plane with upward pointing normals (and given
orientation that is our setting) then
��
��
�
�
�M, N, P� · n d(SA) =
− Mgx − Ngy + P dA.
G
R
Therefore we have
��
�
�
(∇ × x, x + y3 , x2 + y2 − z ) · n d(SA)
G
��
=
�2y, −2x, 1� · n d(SA)
��G
�
�
=
− 2y(2x) − (−2x)(−2y) + (1)(1) dA
��R
�
�
=
− 8xy + 1 dA
R
This integral is easy to calculate. The first part (i.e.,
the −8xy) is 0 by symmetry and the second part (i.e.,
the 1) is the area which is π. We conclude
�
�
�
x, x + y3 , x2 + y2 − z · T ds = π.
∂G