Math 52 - Winter 2007 - Midterm Exam II
Problem 1. (12 pts.) Let Cr be the upper semicircle x2 + y 2 = r2 with y ≥ 0 oriented
from left to right. Let f (x) be any continuous function with a continuous derivative. Show
that
Z
1
(y · f 0 (x)) dx + (x + f (x)) dy = − πr2
2
Cr
Solution: Consider the the half disk D x2 + y 2 ≤ r2 with y ≥ 0. It boundary consists of
the curve −Cr (oriented from right to left) and an interval I that may be parametrized as
(t, 0) for −r ≤ t ≤ r. Then by Green’s Theorem:
Z
Z
0
(y · f 0 (x)) dx + (x + f (x)) dy =
−
(y · f (x)) dx + (x + f (x)) dy +
I
Cr
ZZ
∂
∂
(y · f 0 (x)) +
(x + f (x)) =
−
∂y
∂x
ZZ
D
1
dx dy = πr2
2
D
Using the parametrization (t, 0) of the interval I we get that dy = 0 and y = 0, thus:
Z r
Z
0
0 · f 0 (x) dx = 0
(y · f (x)) dx + (x + f (x)) dy =
−r
I
Thus:
Z
−
1
(y · f 0 (x)) dx + (x + f (x)) dy = πr2
2
Cr
which is the required result.
2
Problem 2. (12 pts.) Let C be the curve parametrized by r(t) = (t·et , sin (2t6 − t3 ) , arctan (t3 ))
for 0 ≤ t ≤ 1. Find
Z
y dx + x dy.
C
Hint: Use the Fundamental Theorem for Line Integrals.
Solution: Hint suggests to look for a potential of hy, xi. It is easily done by PDE’s:
∂
f
∂x
∂
f
∂y
= y =⇒ f (x, y) = xy + C(y)
= x + C 0 (y) = x =⇒ f (x, y) = xy
Now, by Fundamental Theorem for Line Integrals
Z
y dx + x dy = f (B) − f (A)
C
1
where A = r(0) = (0, 0, 0) and B = r(1) = (e, sin 1, arctan 1) are the end points of the curve
C. Thus:
Z
y dx + x dy = e · sin 1
C
Problem 3. (12 pts.) Compute the coordinates of the centroid of the curve that is the
boundary of the triangle with the vertices (0, 0), (2, 0) and (0, 5).
Solution:
From geometry the total mass is the perimeter of the triangle, which is 2 +
√
5 + 22 + 52 . It is easy to find that the diagonal of the triangle is included in the line
y = − 52 x + 5. Thus the perimeter, whihc consists of three curves, may be parametrized as
follows:
C1 :
(t, 0) for 0 ≤ t ≤ 2 =⇒ ds = dt
C2 :
(0, t) for 0 ≤ t ≤ 5 =⇒ ds = dt
√
C3 :
(t, − 52 t + 5) for 0 ≤ t ≤ 2 =⇒ ds = 229 dt
Thus
√
2
Z
m · x̄ =
t+
0
and
5
Z
m · ȳ =
2
t dt +
0
Hence
Z
0
√
29
t dt = 2 + 29
2
√
√
29
25 5 29
5
dt =
+
− t+5
2
2
2
2
√
√
2 + 29
25 + 5 29
√
√
x̄ =
and ȳ =
7 + 29
2(7 + 29)
→
Problem 4. (12 pts.) Show that there is no vector field F such that
→
∇ × F = hx, 0, 0i
→
Solution: Since ∇ ◦ ∇ × F = 0, then:
0 = ∇ ◦ hx, 0, 0i = 1
which is a contradiction.
Problem 5. Fix two positive numbers a and b. Let C be the curve in the xy-plane consisting
of two line segments which join (0, 1) to (a, 0) and then (a, 0) to (a, b).
2
(a) (6 pts.) Evaluate
Z
C
2x dx + 2y dy
x2 + y 2
(the answer will depend on a and b).
Solution: First parametrize both pieces of the curve:
(t, − at + 1) for 0 ≤ t ≤ a =⇒ ds =
(a, t) for 0 ≤ t ≤ b =⇒ ds = dt
C1 :
C2 :
Then
Z
C1
2x dx + 2y dy
=
x2 + y 2
Z
a
√
1+a2
dt
a
2 !a
· −1
t
2
= ln a2
a dt = ln t + 1 −
t 2
2
a
t + 1− a
0
2t + 2 1 −
0
t
a
and
Z
C2
2x dx + 2y dy
=
x2 + y 2
Z
0
b
2t
2
2 b
2
2
dt
=
ln
a
+
t
=
ln
a
+
b
− ln a2
0
2
2
a +t
Thus f (a, b) = ln (a2 + b2 ).
(b) (6 pts.) Is the vector field
→
F =h
x2
2y
2x
, 2
i
2
+ y x + y2
conservative?
→
→
→
Notice: F is not defined at the origin, so saying “∇ × F = 0 ” IS NOT ENOUGH.
→
Solution: The function found in part a) is the only candidate for a potential of F .
Direct check shows that
∇f = h
→
2x
2y
,
i
=
F
x2 + y 2 x2 + y 2
→
thus F is conservative.
3
Problem 6. (12 pts.) Find the value of the integral
ZZ
xyz dS
S
over the surface that is the boundary of the rectangular solid cut from the first octant by
the planes x = a, y = b and z = c.
Solution:
Z Z The surface S has six faces. On three of them one of the coordinates is zero,
xyz dS = 0.
hence
F
The face x = a we can parametrize using 0 ≤ y ≤ b, 0 ≤ z ≤ c. Then the parametrization
is (a, y, z), thus dS = dy dz, and
b
c
ZZ
Z b Z c
1 2 1 2 ab2 c2
xyz dS =
ayz dz dy = a · y · z =
2 0 2 0
4
0
0
F
Similarly, over y = b face:
ZZ
Z
a
Z
c
xyz dS =
0
F
bxz dz dx =
a2 bc2
4
cxy dy dx =
a2 b 2 c
4
0
and lastly, over z = c:
Z
ZZ
a
Z
b
xyz dS =
0
F
Thus
ZZ
xyz dS =
0
abc
(ab + ac + bc)
4
S
Problem 7. Let Γ be a curve parametrized by
γ(t) = (t, 0, f (t))
for some smooth function f (t) and 0 < a ≤ t ≤ b.
(a) (6 pts.) Parametrize the surface S obtained by rotating the curve Γ about the z axis.
Solution:
(t cos θ, t sin θ, f (t))
for 0 ≤ θ ≤ 2π
(b) (4 pts.) Indicate the order of variables in the above parametrization that is needed for
S to be oriented with the normal vector pointing up.
4
Solution: Choose the order randomly and check if it is correct:
→
→
→
0
0
0
N t,θ = T t × T θ = hcos θ, sin θ, f (t)i × h−t sin θ, t cos θ, 0i = h−f (t) · t cos θ, −f (t) · t sin θ, ti
Now
→
→
N t,θ ◦ k = t
so with the above order of parameters the surface is oriented with the normal pointing up.
(as t > 0).
(c) (6 pts.) Setup, but DO NOT EVALUATE an integral representing the surface area of
S.
Solution:
ZZ
Z
b
Z
dS =
S
a
2π
Z b q
q
2
0
t (f (t)) + 1 dθ dt = 2π
t (f 0 (t))2 + 1 dt
0
a
Problem 8. Let S be the open cylinder of radius a, along the x-axis, and bounded by the
two planes x = 0 and x = b. Here “open” means that we do not consider the two end disks
as part of the cylinder.
5
1. (6 pts.) Find a surface parametrization X for S, for which the normal vector points
outwards.
Solution: One parametrization for the cylinder is for instance
X1 (s, t) = (s, a cos t, a sin t),
0 ≤ s ≤ b,
0 ≤ t ≤ 2π.
In this case,
N1 (s, t) =
∂X1 ∂X1
×
(s, t) = (1, 0, 0) × (0, −a sin t, a cos t) = (0, −a cos t, −a sin t).
∂s
∂t
Since both components are negative when t = π/4, for instance, we see that N points
inwards. For it to point outwards, we can swap the variables s and t:
X2 (s, t) = (t, a cos s, a sin s),
0 ≤ t ≤ b,
0 ≤ s ≤ 2π.
Then
N2 (s, t) =
∂X1 ∂X1
∂X2 ∂X2
×
(s, t) = −
×
(t, s) = (0, a cos s, a sin s).
∂s
∂t
∂s
∂t
So the desired parametrization—at least one of them—is X2 . Note that there are other
ways of modifying X1 to make the normal vector pointing outwards; what I have seen
while grading the exam includes 1) swapping sin and cos, or 2) using −s in place of s
for the first variable, or 3) using (s, −a cos t, −a sin t). Note that putting a minus sign
in front of every variable (and modify the (s, t) domain accordingly) does not change
the orientation of the normal (why?).
→
RR
F · dS, where S is oriented such that the normal vector points
→
outwards, and F (x, y, z) = (log(x + 1), y, z).
2. (6 pts.) Compute
S
Solution: I’ll use X2 .
ZZ
Z
F · dS =
S
2π
b
Z
(log(t + 1), a cos s, a sin s) · (0, a cos s, a sin s)dtds
0
Z
= 2π
0
b
(a2 cos2 s + a2 sin2 s)ds = 2πba2 .
0
6