# Difficult Surface Integrals

1. May 23, 2004

### Theelectricchild

I expected Stokes theorem to make my life easier but these problems are even harder than the normal ones ive 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,

Use Stokes' Theorem to evaluate $$\int\int_ScurlFdS$$

where $$F(x, y, z)=< e^{xy}*cos(z) ,x^2*z, xy >$$

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!)

Last edited: May 23, 2004
2. May 23, 2004

### arildno

$$\iint_{S}\nabla\times\vec{F}\cdot{d}\vec{S}=\oint_{C}\vec{F}\cdot{d}\vec{r}$$

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

3. May 23, 2004

### Theelectricchild

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

4. May 23, 2004

### arildno

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

5. May 23, 2004

### arildno

Sorry, first glance wrong..

Last edited: May 23, 2004
6. May 23, 2004

### arildno

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

7. May 23, 2004

thank you :)