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Exam 3 Solutions
APPM 2350, Calculus 3, Fall 2008
November 20, 2008
- Answers in table form:
a. b. c. d. e. f. g. h.
C C C C A C C D
(a) The upper and lower bounding surfaces are hemispheres and this integral com-
putes the volume of a sphere with radius 1. C. 4π/3 is that volume.
(b) Mass is density times area, M = δπ/2, moment about the x axis is
Mx = δ
− 1
∫ √ 1 −x 2
0
y dy dx =
δ
2
− 1
(1 − x^2 ) dx =
δ (1)
The center of mass coordinate is ¯y = Mx/M = (^34) π , so C.
(c) A linear Taylor approximation of f (x, y, z) about (0, 0 , 0) is of the form
L(x, y, z) = f (0, 0 , 0) + fx(0, 0 , 0)x + fy(0, 0 , 0)y + fz (0, 0 , 0)z, (2)
So it has 4 terms. Answer C.
(d) The following algebra shows the answer is a sphere of radius 1/2 centered at
(0, 0 , 1 /2):
ρ = cos φ (multiply both sides by ρ.) ρ^2 = ρ cos φ (substitute z = ρ cos φ
and ρ^2 = x^2 + y^2 + z^2 .)
x^2 + y^2 + z^2 = z (subtract z.) x^2 + y^2 + z^2 − z = 0 (add 1/4.)
x^2 + y^2 + z^2 − z + 1/ 4 = 1 / 4 (factor.) x^2 + y^2 + (z − 1 /2)^2 = (1/2)^2 (this is a sphere, pick C.)
(e) These formulas are on pages 844 and 1044, pick A.
(f) C is a Flux integral, and the only answer not in the box on page 1070.
(g) To make the correct choice, it suffices to choose any pyramid (with any shaped
base) and evaluate its Volume. A particularly simple choice would be the pyra- mid with corners (0, 0 , 0), (1, 0 , 0), (0, 1 , 0), and (0, 0 , 1). This pyramid is the
region between planes x = 0, y = 0, z = 0, and x + y + z = 1. Taking z = 0 as the base and (0, 0 , 1) as the top, we have h = 1 and A = 1/2. Te volume is
∫ (^1)
0
∫ (^1) −x
0
∫ (^1) −x−y
0
dz dy dz =
This agrees with option C.
(h) A conservative field, F = M i + N j + P k, must satisfy Mz = Px, My = Nx, and Nz = Py. For B and C, Mz 6 = Px, so B, and C are eliminated. Also, for A,
Nz 6 = Py, so it can’t be A. D satisfies all three requirements.
- (a) The region is a regular cylinder of radius 1, height 1, with the unit circle in the
xy plane as its base.
(b) ∫ (^2) π
0
0
0
f (r cos θ, r sin θ, z)r dr dz dθ (5)
(c) ∫ (^1)
0
− 1
1 −y^2
−
1 −y^2
f (x, y, z) dx dy dz (6)
(d) z = 1 =⇒ ρ = sec φ and r = 1 =⇒ ρ = csc φ
∫ (^2) π
0
∫ π 4
0
∫ (^) sec φ
0
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ
2 sin φ dρ dφ dθ
∫ (^2) π
0
∫ π 2
π 4
∫ (^) csc φ
0
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ^2 sin φ dρ dφ dθ
(e) z = 1 =⇒ φ = cos−1 1 ρ and r = 1 =⇒ ρ = sin−1 1 ρ
∫ (^2) π
0
0
∫ π 2
0
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ
2 sin φ dφ dρ dθ
∫ (^2) π
0
1
∫ (^) sin−1 1 ρ
cos−1 1 ρ
f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ^2 sin φ dφ dρ dθ
- (a) { C 1 : r 1 (t) =
2 cos t i +
2 sin t j −
π 4 ≤^ t^ ≤^
3 4 π C 2 : r 1 (u) = (2u − 1) i + (1 − 2 u) j 0 ≤ u ≤ 1
(b) F = x i + y j is conservative (My = 0 = Nx) , the circulation of a closed loop is zero by the Fundamental Theorem of Line Integrals.
(c) We need the speed and outward unit normals to build our integrals.
|v 1 (t)| = | −
2 sin t i +
2 cos t j| =
2 , |v 2 (u)| = | 2 i − 2 j| =
n 1 (t) = cos t i + sin t j, n 2 (u) =
(i + j). (13)
C 1
F · n ds =
∫ (^3) π/ 4
−π/ 4
2 cos t i +
2 sin t j) · (cos t i + sin t j)
2 dt
∫ (^3) π/ 4
−π/ 4
2 dt = 2 π
C 2
F · n ds =
0
((2u − 1) i + (1 − 2 u) j) ·
(i + j)
8 dt
0
8 dt = 0
So, total flux is 2π + 0 = 2π.
(d) Convert to polar:
∫ (^3) π/ 4
−π/ 4
0
er
2 rdr dθ =
∫ (^3) π/ 4
−π/ 4
(e
√ (^2) − 1) dθ
π
2
(e
√ (^2) − 1).
Some alternative solution options for Problem 4:
4a. We can parametrize the two boundary segments as
C 1 , and C 2 ,.
x = 2 cos t
y = 2 sin t
[ t [
x = t
y = − t
− 1 [ t [ 1
b. Using the second formulation of the integral, we start by noting that the parametrizations
above imply
C 1 , and C 2 ,.
dx = − 2 sin t dt
dy = 2 cot t dt
[ t [
dx = dt
dy = − dt
− 1 [ t [ 1
Therefore, with M = x and N = y
¶ C
1
M dx + N dy = ¶−/
3 /
(( 2 cos t )(− 2 sin t ) + ( 2 sin t )( 2 cos t )) dt = ¶−/
3 /
0 dt = 0
C 2
M dx + N dy = ¶
− 1
1
t dt − ¶
− 1
1
− t dt = 0 + 0 = 0
giving
C
C 1
C 2
c. With the same parametrization:
C 1
M dy − N dx = ¶
−/
3 /
(( 2 cos t )( 2 cos t ) − ( 2 sin t )(− 2 sin t )) dt = 2 ¶
−/
3 /
1 dt = 2
C 2
M dy − N dx = ¶
− 1
1
− t dt − ¶
− 1
1
− t dt = 0 + 0 = 0
giving
¶ C = ¶ C
1
+ ¶ C
2
In still another approach, we can use Green's theorem. Although this is not part of the test, it is
still allowed to be used. Then:
4 b. ¶.
C
M dx + N dy = ¶¶
Ø N
Ø x −^
Ø M
Ø y dxdy^ =^ ¶¶(0^ +^ 0) dxdy^ =^0
4 c. ¶ C M dy − N dx = ¶¶.
Ø M
Ø x +^
Ø N
Ø y dxdy^ =^ ¶¶
(1 + 1) dxdy = 2 & {area of region} = 2
