Math 174  Extra Credit  D. Gottlieb

Peter Schlamiel tried to use Stokes’ theorem

∮_{C} F⃗ · dr⃗ = ∫∫_{S}(∇ × F⃗) · n⃗ dσ

where S is the surface {(x, y, z) so that x^{2} + y^{2} + 4z^{2} = 4 and y ≤ 1} and C is the boundary closed curve on the surface given by y = 1. Here F⃗ = \frac{x}{x^{2} + y^{2}} i⃗ + \frac{y}{x^{2} + y^{2}} j⃗ + xk⃗.

Find Schlamiel’s mistakes. One point for each error you find, minus one point for each correct thing you say is an error. Give reasons for your statements. Unclear or incorrect reasons will receive 0 or −1 points. Correct reasons will get +1 points each:

∇ × F⃗ = −j⃗
n⃗ = 2xi⃗ + 2yj⃗ + 8zk⃗
(∇ × F⃗) · n⃗ = −2y.

Parameterize S by r⃗(u, v) = v cos ui⃗ + v sin uj⃗ + \frac{1}{2} \sqrt{4 − v^{2}} k⃗ where 0 ≤ u ≤ 2π and −2 ≤ v ≤ 2.

r⃗_{u} = −v sin ui⃗ + v cos uj⃗
r⃗_{v} = cos ui⃗ + sin uj⃗ − \frac{v}{2\sqrt{4 − v^{2}}} k⃗
r⃗_{u} × r⃗_{v} = \frac{−v^{2} cos u}{2\sqrt{4 − v^{2}}} i⃗ + (\frac{−v^{2} sin u}{2\sqrt{4 − v^{2}}} ) j⃗ − vk⃗

dσ = \sqrt{\frac{v^{4}}{4(4 − v^{2})} + v^{2}} dudv

∫∫_{S}(∇ × F⃗) · n⃗dσ = ∫_{−2}^{2} ∫_{0}^{2π}(−2v cos u)\sqrt{\frac{v^{4}}{16 − 4v^{2}} + v^{2}} du dv

Now C = {(x, y, z)|x^{2} + 1 + 4z^{2} = 4, y = 1}. Let r⃗(x) parametrize C where

r⃗(x) = xi⃗ + j⃗ + \frac{1}{2}\sqrt{3 − x^{2}} k⃗, where − \sqrt{3} ≤ x ≤ \sqrt{3}

\frac{dr⃗}{dx} = (i⃗ − \frac{x}{2\sqrt{3 − x^{2}}} k⃗)
F⃗ · \frac{dr⃗}{dx} = (\frac{x}{x^{2} + 1} i + \frac{1}{x^{2} + 1} j + xk⃗) · (i⃗ − \frac{x}{2\sqrt{3 − x^{2}}} k)
F⃗ · \frac{dr}{dx} = \frac{x}{x^{2} + 1} − \frac{x^{2}}{2\sqrt{3 − x^{2}}}

∮_{C} F⃗ · dr⃗ = ∫_{−\sqrt{3}}^{√\overline{3}} (\frac{x}{x^{2} + 1} − \frac{x^{2}}{2\sqrt{3 − x^{2}}} ) dx

So

∫_{−2}^{2} ∫_{0}^{2π}(−2v cos u)\sqrt{\frac{v^{4}}{16 − 4v^{2}} + v^{2}} dv du

= ∫_{−\sqrt{3}}^{√\overline{3}} (\frac{x}{x^{2} + 1} − \frac{x^{2}}{2\sqrt{3} − x^{2}} ) dx