Calculus II, Spring 2014

Brief Solutions to Final Exam

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1. (15+1 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.

2. (15 pts) Let D = {1 ≤ x ≤ 2, 0 ≤ xy ≤ 1, 0 ≤ z ≤ 1}. Evaluate

∫ ∫ ∫_{D}(x^{2}y + xyz)dV.

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

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

(b) Prove either one of the two forms on R (20 pts). (10 pts) If you can not prove it for general F, do the following: 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.

Ans:

(a) Let C be the boundary of R.

∳_{C} F · n ds = ∳_{C} M dy − N dx = ∬_{R} (\frac{∂M}{∂x} + \frac{∂N}{∂y} ) dx dy,

∳_{C} F · T ds = ∳_{C} M dx + N dy = ∬_{R} (\frac{∂N}{∂x} − \frac{∂M}{∂y} ) dx dy.

(b1) (Prove of Green’s Theorem, for instance, in the first form.) Let C = C_{1} ∪C_{2} ∪C_{3}, where

\begin{matrix}C_{1} : & r(t) = (\sqrt{1 − t^{2}}, t) , & 0 ≤ t ≤ 1, \\ C_{2} : & r(t) = (0, 1 − t) , & 0 ≤ t ≤ 1, \\ C_{3} : & r(t) = (t, 0) , & 0 ≤ t ≤ 1. \end{matrix}

Then (changing of variables u = 1 − t, v = \sqrt{1 − t^{2}})

∫_{C_{1}} M dy = ∫_{0}^{1} M(\sqrt{1 − t^{2}}, t) · 1 dt = ∫_{0}^{1} M(\sqrt{1 − t^{2}}, t) dt,
∫_{C_{2}} M dy = ∫_{0}^{1} M(0, 1 − t) · (−1) dt = − ∫_{0}^{1} M(0, u) du,
∫_{C_{3}} M dy = ∫_{0}^{1} M(t, 0) · 0 dt = 0,
∫_{C_{1}} N dx = ∫_{0}^{1} N(\sqrt{1 − t^{2}}, t) · \frac{d \sqrt{1 − t^{2}}}{dt} dt = − ∫_{0}^{1} N(v, \sqrt{1 − v^{2}}) dv,
∫_{C_{2}} N dx = ∫_{0}^{1} N(0, 1 − t) · 0 dt = 0,
∫_{C_{3}} N dx = ∫_{0}^{1} N(t, 0) · 1 dt = ∫_{0}^{1} N(t, 0) dt.

Hence,

∫_{C} M dy − N dx

= (∫_{C_{1}} M dy + ∫_{C_{2}} M dy + ∫_{C_{3}} M dy) − (∫_{C_{1}} N dx + ∫_{C_{2}} N dx + ∫_{C_{3}} N dx)

= ∫_{0}^{1} M(\sqrt{1 − t^{2}}, t) dt − ∫_{0}^{1} M(0, u) du + ∫_{0}^{1} N(v, \sqrt{1 − v^{2}}) dv − ∫_{0}^{1} N(t, 0) dt. On the other hand,

∬_{R} (\frac{∂M}{∂x} + \frac{∂N}{∂y} ) dx dy

= ∬_{R} \frac{∂M}{∂x} dx dy + ∬_{R} \frac{∂N}{∂y} dx dy

= ∫_{0}^{1} ∫_{0}^{√\overline{1−y^{2}}} \frac{∂M}{∂x} dx dy + ∫_{0}^{1} ∫_{0}^{√\overline{1−x^{2}}} \frac{∂N}{∂y} dy dx

= ∫_{0}^{1} [M(x, y)]_{x=0}^{x=\sqrt{1−y^{2}}} dy + ∫_{0}^{1} [N(x, y)]_{y=0}^{y=\sqrt{1−x^{2}}} dx

= ∫_{0}^{1} (M(\sqrt{1 − y^{2}}, y) − M(0, y)) dy + ∫_{0}^{1} (N(x, \sqrt{1 − x^{2}}) − N(x, 0)) dx. Hence,

∫_{C} M dy − N dx = ∬_{R} (\frac{∂M}{∂x} + \frac{∂N}{∂y} ) dx dy.

(b2) (Verifying the case F = (y, x).) Let C = C_{1} ∪ C_{2} ∪ C_{3}, where

\begin{matrix}C_{1} : & r(t) = (cos t, sin t) , & 0 ≤ t ≤ \frac{π}{2}, \\ C_{2} : & r(t) = (0, 1 − t) , & 0 ≤ t ≤ 1, \\ C_{3} : & r(t) = (t, 0) , & 0 ≤ t ≤ 1. \end{matrix}

We have

∳_{C} y dy − x dx = ∫0^{\frac{π}{2}} sin t cos t + cos t sin t dt + ∫_{0}^{1} (1 − t) (−1) dt + ∫_{0}^{1} (−t) dt = 0,

∬_{R} (\frac{∂M}{∂x} + \frac{∂N}{∂y} ) dx dy = ∬_{R} 0 dx dy = 0,

and

∳_{C} x dx + y dy = ∫0^{\frac{π}{2}} cos t (− sin t) + sin t cos t dt + ∫_{0}^{1} 0 dt + ∫_{0}^{1} 0 dt = 0,

∬_{R} (\frac{∂N}{∂x} − \frac{∂M}{∂y} ) dx dy = ∬_{R} 0 dx dy = 0.

4. (15 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 · T ds depends only on f, (x(0), y(0), z(0)) and (x(1), y(1), z(1)). Ans: True. Let g(t) = f(x(t), y(t), z(t)). Then

\frac{dg}{dt} = \frac{∂f}{∂x} \frac{dx}{dt} + \frac{∂f}{∂y} \frac{dy}{dt} + \frac{∂f}{∂z} \frac{dz}{dt} = ∇f · \frac{∂r}{∂t} .

Hence

∫_{C} ∇f · T ds = ∫_{C} ∇f · dr = ∫_{0}^{1} ∇f · \frac{dr}{dt} dt = ∫_{0}^{1} \frac{dg}{dt} dt = [g(t)]_{t=0}^{t=1}
= g(1) − g(0) = f(x(1), y(1), z(1)) − f(x(0), y(0), z(0)).

5. (30 pts) Let R = {1/4 ≤ x^{2} + y^{2} ≤ 4}, F (x, y) = (2y, x), G(x, y) = (x, y), 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. Ans: (4% for answer and 6% for reason each)

(a) Since 2 = \frac{∂}{∂y} (2y) ≠ \frac{∂}{∂x} (x) = 1. Hence, F is NOT conservative.

(b) Since G = ∇g, where g = \frac{1}{2} (x^{2} + y^{2}). Hence, G is conservative.

(c) Let C : r(t) = (cos t, sin t), 0 ≤ t ≤ 2π. Then

∫_{C} H · dr = ∫_{0}^{2π} (− sin t, cos t) · (− sin t, cos t) dt = 2π.

Hence, H is NOT conservative.

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 _{∫ ∫S}F · n dσ where n is unit normal of S pointing away from the origin. Ans: We parametrize the surface S by

r(θ, φ) = (2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ) , 0 ≤ θ ≤ \frac{π}{2} , 0 ≤ φ ≤ \frac{π}{2} .  (3%)

Then r_{θ} = (−2 sin φ sin θ, 2 sin φ cos θ, 0), r_{φ} = (2 cos φ cos θ, 2 cos φ sin θ, −2 sin φ), and

r_{θ} × r_{φ} = |\begin{matrix}i & j & k \\ −2 sin φ sin θ & 2 sin φ cos θ & 0 \\ 2 cos φ cos θ & 2 cos φ sin θ & −2 sin φ \end{matrix}|
= (−4 sin^{2} φ cos θ, −4 sin^{2} φ sin θ, −4 sin φ cos φ) .  (6%)

Take

n = −\frac{r_{θ} × r_{φ}}{∥r_{θ} × r_{φ}∥} = \frac{(4 sin^{2} φ cos θ, 4 sin^{2} φ sin θ, 4 sin φ cos φ)}{∥r_{θ} × r_{φ}∥}.  (1%)

and we have dσ = ∥r_{θ} × r_{φ}∥ dθ dφ,

F · n = (0, 0, 2 cos φ) · \frac{(4 sin^{2} φ cos θ, 4 sin^{2} φ sin θ, 4 sin φ cos φ)}{∥r_{θ} × r_{φ}∥} = \frac{8 sin φ cos^{2} φ}{∥r_{θ} × r_{φ}∥}.

Hence,

∬_{S} F · n dσ = ∫0^{\frac{π}{2}} ∫0^{\frac{π}{2}} \frac{8 sin φ cos^{2} φ}{∥r_{θ} × r_{φ}∥} · ∥r_{θ} × r_{φ}∥ dθ dφ
= 8 ∫0^{\frac{π}{2}} cos^{2} φ sin φ dφ ∫0^{\frac{π}{2}} dθ = 8[ − \frac{1}{3} cos^{3} φ]_{φ=0}^{φ= \frac{π}{2}} [θ]_{θ=0}^{θ= \frac{π}{2}} = \frac{4π}{3} . (5%)