# Homework Help: Stokes' Theorem problem

1. Mar 20, 2010

### mathman44

1. The problem statement, all variables and given/known data

Use stokes' theorem to find I = $$\int\int (\nabla x F) n dS$$ where D is the part of the sphere $$x^2 + y^2 + (z-2)^2 = 8$$ that lies above the xy plane, and

$$F=ycos(3xz^2)i + x^3e^[-yz]j - e^[zsinxy]k$$

Attempt at solution:
I want to use the line integral $$\int F dr$$ to solve this.

I parametrize the boundary $$r(t) = (2cos\theta)i + (2sin\theta)j + 0k)$$ (the sphere on the xy plane)

Then F dotted with r'[t] is $$\int -4(sin\theta)^2 + (2cos\theta)^4$$

and 0 < theta < 2pi

And the answer that I get is 8pi. Is this correct? More specifically is my boundary curve legit, considering that the sphere "bulges" out above this boundary? :S

Last edited: Mar 21, 2010
2. Mar 20, 2010

### mathman44

I just realized that on the xy plane the radius would be 2, not sqrt(8). Nonetheless, is this otherwise okay?

3. Mar 20, 2010

### ideasrule

This is not what I got. Did you get dr/d(theta)=-2sin(theta)i + 2cos(theta)j?

That's not a problem. The boundary can curve in or out as much as it wants, but Stokes' theorem always holds true.

4. Mar 20, 2010

### mmmboh

Is the answer you got 0?

5. Mar 21, 2010

### ideasrule

Me? Yes, I got 0.

6. Mar 21, 2010

### Dustinsfl

Since stoke's theorem is an alt of green's, wouldn't it be easier to just compare the 6 partial derivatives? Because the integral will be 0 if it is conservative which occurs if they are all equal, correct?

7. Mar 21, 2010

### mathman44

I changed the r[t] to $$2cos\theta,2sin\theta,0$$. For the F dot r'[t] I got $$\int -4(sin\theta)^2 + (2cos\theta)^4$$ with theta from 0 to 2pi. Why is this not giving me zero :S ?

8. Mar 21, 2010

### ideasrule

Try it. I don't think the field is conservative.

9. Mar 21, 2010

### ideasrule

You've done everything right so far. If you don't get 0, it's just a mistake in your integration somewhere.

10. Mar 21, 2010

### mathman44

I double checked on maple and 8pi seems correct, but I get zero when I directly compute the double integral without using Stokes'...

Last edited: Mar 21, 2010
11. Mar 21, 2010

### gabbagabbahey

I get $8\pi$ using both methods, so I'm pretty sure you are just doing something wrong when you calculate the double integral directly.

12. Mar 21, 2010

### mathman44

Can anyone confirm this? We have two homework helpers with two different answers!

13. Mar 21, 2010

### mathman44

Shameless bump

14. Mar 21, 2010

### gabbagabbahey

I've asked in the homework helpers forum for another HH to take a look when they get time, so hopefully someone will reply.

15. Mar 22, 2010

### vela

Staff Emeritus
I get $8\pi$ for the line integral. Is there a trick to evaluating the double integral? I get a mess for curl F.

16. Mar 22, 2010

### mmmboh

When I found the curl I took z to equal 0, and so the curl ends up being 3x2-1.
I then did $$\int\int3x^{2}-1 dA = \int_{0}^{2\pi}\int_{0}^{2}(r^{3}cos^2 {\phi} - r) drd{\phi}$$
this gives me zero. There is a similar problem in my book and this seems to be how it's done...

17. Mar 22, 2010

### ideasrule

Are we all integrating $$\int -4(sin\theta)^2 + (2cos\theta)^4$$ from 0 to 2pi?

18. Mar 22, 2010

### mmmboh

I did my integral a different way.

19. Mar 22, 2010

### ideasrule

Which way?

20. Mar 22, 2010

### mmmboh

I took the curl on the cirlce in the xy plane, so z=0.

21. Mar 22, 2010

### vela

Staff Emeritus
That's not right. You have to integrate over the surface of the sphere, so z won't be zero. Also, you apparently set z=0 before you took the curl of F, which you shouldn't do either.

22. Mar 22, 2010

### vela

Staff Emeritus
Yeah, gabba, mmmboh, and I all get 8 pi for that integral.

23. Mar 22, 2010

### gabbagabbahey

Two things:

(1) You seem to be integrating over the flat disk that is bounded by the same circle that the OPs original surface is bounded by. This should give the same result, but is not really a direct computation of the original double integral

(2) What happened to the factor of 3? ....[itex](3x^2-1)dA=(3r^{3}cos^2 {\phi} - r) dr d\phi\neq (r^{3}cos^2 {\phi} - r) dr d\phi[/tex]

24. Mar 22, 2010

### gabbagabbahey

By an odd coincidence, you actually get the same result either way.

25. Mar 22, 2010

### mmmboh

Ah you're right, I actually do have the factor 3 written down, but I lost it somewhere...I fixed my work and I get $$8\pi$$