Calculus II, Spring 2016

Brief answers to Final Exam

Jun 16, 2016

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1. (10 pts) Replace

∫ _{0}^{2π} ∫ _{0}^{1} ∫ _{0}^{√ \overline{2−r^{2}}} rdz dr dθ

by triple integrals in spherical coordinates in the order dρdϕdθ and dϕdρdθ, respectively. Need not evaluate them. Answer.

∫ _{0}^{2π} [∫ 0^{\frac{π}{4}} ∫ _{0}^{√ \overline{2}} ρ^{2} sin ϕ dρdϕ + ∫ \frac{π}{2} ; \frac{π}{4} ∫ _{0}^{cscϕ} ρ^{2} sin ϕ dρdϕ] dθ (5 pts)

∫ _{0}^{2π} [∫ _{0}^{1} ∫ 0^{\frac{π}{2}} ρ^{2} sin ϕ dϕdρ + ∫ _{1}^{√ \overline{2}} ∫ _{0}^{csc^{−1} ρ} ρ^{2} sin ϕ dϕdρ] dθ (5 pts)

2. (15 pts) Evaluate the surface area of S = \{z = \sqrt{x^{2} + y^{2}} , 1 ≤ xy ≤ 2, 1 ≤ x/y ≤ 3, x > 0, y > 0}. Answer.

Method.1 Let f (x, y) = \sqrt{x^{2} + y^{2}} . First compute

f_{x} = \frac{x}{\sqrt{x^{2} + y^{2}}} , f_{y} = \frac{y}{\sqrt{x^{2} + y^{2}}} . (2 pts)

Then the surface area is

Area = _{R} \sqrt{f_{x}^{2} + f_{y}^{2} + 1} dxdy (2pts) = _{R} \sqrt{2}dxdy. (1pt)

Let u = xy, v = \frac{x}{y}, x, y > 0 (2 pts). Then 1 ≤ u ≤ 2, 1 ≤ v ≤ 3 (2 pts), and the Jacobian is

| \frac{1}{2 \sqrt{uv}} −\frac{1}{2} \frac{\sqrt{u}}{v \sqrt{v}}^{\frac{\sqrt{v}}{2 \sqrt{u}} \frac{\sqrt{u}}{2 \sqrt{v}}} | = −\frac{1}{2v}. (2 pts)

Thus, The integral is changed into

Area = ∫ _{1}^{2} ∫ _{1}^{3} \frac{\sqrt{2}}{2v} dvdu (2 pts) = \frac{ln 3}{\sqrt{2}}. (2 pts)

Method.2 Let r(r, θ) = (r cos θ, r sin θ, r), (\frac{2}{sin 2θ})^{\frac{1}{2}} ≤ r ≤ (\frac{4}{sin 2θ})^{\frac{1}{2}}, cot^{−1} 3 ≤ θ ≤ cot^{−1} 1 (3 pts). First compute

r_{r}(r, θ) = (cos θ, sin θ, 1), (1 pt)
r_{θ}(r, θ) = (−r sin θ, r cos θ, 0), (1 pt)
|r_{r}(r, θ) × r_{θ}(r, θ)| = |(−r cos θ, −r sin θ, r)| = \sqrt{2}r. (2 pts)

Thus, the area is

∫ _{cot^{−1} 3}^{cot^{−1} 1} ∫ \begin{matrix}(\frac{4}{sin 2θ}) & \frac{1}{2} \\ (\frac{2}{sin 2θ}) & \frac{1}{2} \end{matrix} \sqrt{2}r drdθ (2 pts) = \sqrt{2}∫ _{cot^{−1} 3}^{cot^{−1} 1} \frac{1}{2} (\frac{4}{sin 2θ} − \frac{2}{sin 2θ} ) dθ (2 pts)

= \sqrt{2}∫ _{cot^{−1} 3}^{cot^{−1} 1} csc2θ dθ
= − \frac{1}{\sqrt{2}} [ln(1 + 0) − ln(\frac{5}{3} + \frac{4}{3})] (2 pts)
= \frac{ln 3}{\sqrt{2}}. (2 pts)

3. Let F(x, y) = (M(x, y), N(x, y)) have continuous first and second derivatives everywhere in ^{2} and let R = {x^{2} + y^{2} < 1, x > 0, y > 0}.

(a) (10 pts) State Green’s Theorem in both forms for Fon R.

(b) (16 pts) Take F = (y, x) and verify both forms on R. That is, evaluate both line integral and double integral and check that they are the same. Do this for both forms.

Answer.

(a)

∮_{C}F· nds = _{C}Mdy − Ndx = _{R}(M_{x} + N_{y}) dxdy (5 pts)
CF· Tds = _{C}Mdx + Ndy = _{R}(N_{x} − M_{y}) dxdy (5 pts)

where C is a piecewise smooth, simple closed curve which encloses the region R.

(b) Let r_{1}(t) = (cos t, sin t), 0 ≤ t ≤ \frac{π}{2}; r_{2}(t) = (0, 1 − t), 0 ≤ t ≤ 1; r_{3}(t) = (t, 0), 0 ≤ t ≤ 1 (4 pts). Then

∮_{C}F· n ds = ∫ 0^{\frac{π}{2}}(sin t cos t − cos t(− sin t)) dt + ∫ _{0}^{1}((1 − t) · (−1) − 0 · 0) dt
+ ∫ _{0}^{1}(0 · 0 − t · 1) dt
= 1 − \frac{1}{2} − \frac{1}{2} = 0 = _{R}(0 + 0) dxdy = _{R}(M_{x} + N_{y}) dxdy, (6 pts)
CF· T ds = ∫ 0^{\frac{π}{2}}(sin t(− sin t) + cos t cos t) dt + ∫ _{0}^{1}((1 − t) · 0 + 0 · (−1)) dt
+ ∫ _{0}^{1}(0 · 1 + t · 0) dt
= 0 + 0 + 0 = _{R}(1 − 1) dxdy = _{R}(N_{x} − M_{y}) dxdy. (6 pts)

4. (10 pts) True or false? Give details.

If f (x, y, z) has continuous first derivatives in a domain D, and C = {(x(t), y(t), z(t)), 0 ≤ t ≤ 1} be a smooth curve in D. Then ∫_{C}∇f · Tds depends only on f, (x(0), y(0), z(0)) and (x(1), y(1), z(1)). Answer. True. (2 pts)

∫_{C}∇f · T ds

= ∫ _{0}^{1}(f_{x}(x(t), y(t), z(t))x′(t) + f_{y}(x(t), y(t), z(t))y′(t) + f_{z}(x(t), y(t), z(t))z′(t)) dt (4 pts)
= ∫ _{0}^{1} ∂_{t}(f (x(t), y(t), z(t)) dt = f (x(1), y(1), z(1)) − f (x(0), y(0), z(0)). (4 pts)

5. (24 pts) Let R = \{1/4 ≤ x^{2} + y^{2} ≤ 4\}, F(x, y) = (2y, x), G(x, y) = ( \frac{x}{\sqrt{x^{2} + y^{2}}} , \frac{y}{\sqrt{x^{2} + y^{2}}} ),

H(x, y) = (\frac{−y}{x^{2} + y^{2}} , \frac{x}{x^{2} + y^{2}} ).

Which one(s) of F, G and H is (are) conservative on R? (That is, which one(s) of ∫_{C}F· dr, ∫_{C}G · dr and ∫_{C}H · dr is (are) zero on \underline{every} closed loop C in R?) Explain. Answer. Fis not conservative (2 pts) since ∂_{y}(2y) = 2 1 = ∂_{x}(x) (2 pts). H is not conservative (2 pts) since for some r(t) = (a cos t, a sin t), 0 ≤ t ≤ 2π (2 pts),

CH · Tds = ∫ _{0}^{2π} (−\frac{sin t}{a}, \frac{cos t}{a}) · (−a sin t, a cos t) dt = ∫ _{0}^{2π} 1 dt = 2π 0. (6 pts)

G is conservative. (2 pts) There are two methods to check it.

Method.1 It suffices to check whether _{C}G· Tds = 0 for any loop of the form C = {x^{2}+ y^{2} = a^{2}, 1/2 ≤ a ≤ 2} since G satisfies the component test.

∂_{y} ( \frac{x}{\sqrt{x^{2} + y^{2}}} ) = −\frac{xy}{(x^{2} + y^{2})^{3/2}} = ∂_{x} ( \frac{y}{\sqrt{x^{2} + y^{2}}} ) (2 pts).

Let r(t) = (a cos t, a sin t), 0 ≤ t ≤ 2π.

CG · Tds = ∫ _{0}^{2π} (cos t, sin t)· (−a sin t, a cos t) dt = ∫ _{0}^{2π} 0 dt = 0. (6 pts)

Method.2 Find f = f (x, y) such that G = ∇f.

f_{x}(x, y) = \frac{x}{\sqrt{x^{2} + y^{2}}}

⇒  f (x, y) = \sqrt{x^{2} + y^{2}} + g(y)
⇒  f_{y}(x, y) = \frac{y}{\sqrt{x^{2} + y^{2}}} + g′(y) = \frac{y}{\sqrt{x^{2} + y^{2}}}
⇒  g(y) = C
⇒  f (x, y) = \sqrt{x^{2} + y^{2}} + C. (8 pts)

6. (15 pts) Let F(x, y, z) = zk and S = {x^{2} + y^{2} + z^{2} = 4, x > 0, y > 0, z > 0}. Evaluate

∫ ∫ SF· n dσ where n is unit normal of S pointing away from the origin.

Answer.

Method.1 Let r(ϕ, θ) = (2 sin ϕ cos θ, 2 sin ϕ sin θ, 2 cos ϕ), 0 ≤ ϕ, θ ≤ \frac{π}{2} (3 pts). Then

r_{ϕ}(ϕ, θ) = (2 cos ϕ cos θ, 2 cos ϕ sin θ, −2 sin ϕ) (1 pt)
r_{θ}(ϕ, θ) = (−2 sin ϕ sin θ, 2 sin ϕ cos θ, 0) (1 pt)
r_{ϕ} × r_{θ} = (4 sin^{2} ϕ cos θ, 4 sin^{2} ϕ sin θ, 4 cos ϕ sin ϕ). (2 pts)

Therefore,

S F·ndσ = ∫ 0^{\frac{π}{2}} ∫ 0^{\frac{π}{2}}(0, 0, 2 cos ϕ)·(4 sin^{2} ϕ cos θ, 4 sin^{2} ϕ sin θ, 4 cos ϕ sin ϕ) dϕdθ (4 pts)

= 8 ∫ 0^{\frac{π}{2}} ∫ 0^{\frac{π}{2}} cos^{2} ϕ sin ϕ dϕdθ = 4π · ∫ _{0}^{1} u^{2} du (2 pts) = \frac{4π}{3} . (2 pts)

Method.2 Let g(x, y, z) = x^{2} + y^{2} + z^{2} = 4. Then

∇g = (2x, 2y, 2z), |∇g · p| = 2z. (7 pts)

RF· ndσ =  _{R}F· \frac{∇g}{|∇g|} \frac{|∇g|}{|∇g · p|}dxdy
=  _{R}F· \frac{∇g}{|∇g · p|}dxdy (4 pts)
=  _{R}z dxdy (2 pts)
= the volume of the sphere of radius 2 in the first octant
= \frac{4π · 2^{3}}{3} · \frac{1}{8}
= \frac{4π}{3} . (2 pts)

7. (10 pts) Let f (x, y) = 2x + 3y + 4 + ( \sqrt{x^{2} + y^{2}} )^{\frac{3}{2}}. Is f (x, y) differentiable at (0, 0)? Explain. Answer. Yes. (2 pts) First we compute

f_{x}(0, 0) = lim_{h→0} \frac{f (h, 0) − f (0, 0)}{h} = lim_{h→0} \frac{2h + h^{3}}{h} = 2 (2 pts)

f_{y}(0, 0) = lim_{h→0} \frac{f (0, h) − f (0, 0)}{h} = lim_{h→0} \frac{3h + h^{3}}{h} = 3. (2 pts)

Given (△x, △y). Then

f (△x, △y) − f (0, 0) = 2△x + 3△y + ( \sqrt{△x^{2} + △y^{2}} )^{\frac{3}{2}} .(2 pts)

To show ( \sqrt{△x^{2} + △y^{2}} )^{\frac{3}{2}} = ϵ_{1}△x + ϵ_{2}△y. (2 pts)

Method.1 Write (△x, △y) = (r cos θ, r sin θ). Then

( \sqrt{△x^{2} + △y^{2}} )^{\frac{3}{2}} = r^{\frac{3}{2}} = r · r^{\frac{1}{2}} = r(ϵ_{1} cos θ + ϵ_{2} sin θ) = ϵ_{1}△x + ϵ_{2}△y
for some ϵ_{1}, ϵ_{2} which approach to zero as △x, △y → 0.

Method.2

( \sqrt{△x^{2} + △y^{2}} )^{\frac{3}{2}} = ( \sqrt{△x^{2} + △y^{2}} )^{\frac{1}{2}} \sqrt{△x^{2} + △y^{2}} = o(1) \sqrt{△x^{2} + △y^{2}} =^{HW7} ϵ_{1}△x+ϵ_{2}△y

for some ϵ_{1}, ϵ_{2} which approach to zero as △x, △y → 0.

8. (10 pts) Evaluate ∑_{n=0}^{∞} \frac{x^{n}}{n + 1} on |x| < 1 using computational rules of power series.

Answer.

For x = 0, clearly ∑ _{n=0}^{∞} \frac{x^{n}}{n+1} = 1. For 0 < |x| < 1,

∑_{n=0}^{∞} x^{n} = \frac{1}{1 − x} (4 pts)

⇒  ∑_{n=0}^{∞} \frac{x^{n+1}}{n + 1} = − ln(1 − x) (4 pts)

⇒  ∑_{n=0}^{∞} \frac{x^{n}}{n + 1} = −\frac{ln(1 − x)}{x}. (2 pts)