Calculating Surface Integrals with Stokes' Theorem

[Theelectricchild]

I expected Stokes theorem to make my life easier but these problems are even harder than the normal ones I've been doing.


Use Stokes' Theorem to evaluate \int\int_ScurlFdS

where F(x, y, z) = < x^2*y^3*z, sin(xyz) ,xyz >

S: Part of cone y^2 = x^2 + z^2 that lies between the planes y = 0 and y = 3 oriented in the direction of the positive y - axis.

not sure about this one... I parameterized the curve C as r(t) = < 3cost, 3, -3sint > to get a curve with positive orientation induced by S having normal vector in pos-y direction...a circle in plane y = 3,
answer: 2187*pi/4

Use Stokes' Theorem to evaluate \int\int_ScurlFdS

where F(x, y, z)=&lt; e^{xy}*cos(z) ,x^2*z, xy &gt;

S: Hemisphere x = sqrt{1-y^2-z^2} oriented in the direction of the positive x - axis.

curve C : r(t) = <0, cost, sint> ... a circle again in plane x=0, hmmm, wonder if I'm messing these up...?
answer: 0 (this seemed way too easy...everything was zero!)

 

[arildno]

I assume you're talking about this verson of Stokes' theorem:
\iint_{S}\nabla\times\vec{F}\cdot{d}\vec{S}=\oint_{C}\vec{F}\cdot{d}\vec{r}

In that case, your last calculation is correct; even if it is indecently easy..

 

[Theelectricchild]

Even for the first one? The answer just seemed completely off considering its size.

 

[arildno]

At a glance, the first one ought to be zero as well..

 

[arildno]

Sorry, first glance wrong..

 

[arildno]

If 3^{7}=2187, then the first should be correct as well.

 

[Theelectricchild]

thank you :)