# Stokes' Theorem problem

mathman44

## Homework Statement

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

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## Answers and Replies

mathman44
I just realized that on the xy plane the radius would be 2, not sqrt(8). Nonetheless, is this otherwise okay?

Homework Helper

## Homework Statement

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 $$r(t) = (\sqrt(8)cos\theta)i + (\sqrt(8)sin\theta)j + 0k)$$ (the sphere on the xy plane)

Then F dotted with r'[t] is $$64(cos\theta)^4 - 8sin\theta$$

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

More specifically is my boundary curve legit, considering that the sphere "bulges" out above this boundary? :S

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

mmmboh
Is the answer you got 0?

Homework Helper
Is the answer you got 0?

Me? Yes, I got 0.

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?

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 ?

Homework Helper
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?

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

Homework Helper
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 ?

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

mathman44
I double checked on maple and 8pi seems correct, but I get zero when I directly compute the double integral without using Stokes'...

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

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.

mathman44
Can anyone confirm this? We have two homework helpers with two different answers!

mathman44
Shameless bump

Homework Helper
Gold Member
Can anyone confirm this? We have two homework helpers with two different answers!

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

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

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...

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

mmmboh
I did my integral a different way.

Homework Helper
Which way?

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...

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

Staff Emeritus
Homework Helper
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.

Staff Emeritus
Homework Helper
Are we all integrating $$\int -4(sin\theta)^2 + (2cos\theta)^4$$ from 0 to 2pi?
Yeah, gabba, mmmboh, and I all get 8 pi for that integral.

Homework Helper
Gold Member
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...

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? ....$(3x^2-1)dA=(3r^{3}cos^2 {\phi} - r) dr d\phi\neq (r^{3}cos^2 {\phi} - r) dr d\phi[/tex] Homework Helper Gold Member Also, you apparently set z=0 before you took the curl of F, which you shouldn't do either. By an odd coincidence, you actually get the same result either way. 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 $$8\pi$$ Homework Helper Gold Member Are we all integrating $$\int -4(sin\theta)^2 + (2cos\theta)^4$$ from 0 to 2pi? $$\int_0^{2\pi}\sin^n\theta d\theta=\int_0^{2\pi}\cos^n\theta d\theta=0$$ is only true for odd [itex]n$....Perhaps this is the source of your error?

Homework Helper
No, the source of my error was just sloppiness. (Specifically, I thought that cos^2 (x)=(cos(2x)-1)/2. It took an amazingly long time to fix that.) Now I get 8*pi, in agreement with everybody else.

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

I don't think it's a coincidence. z doesn't change across the flat surface, so it's a constant, and whether you set a constant before or after integration shouldn't matter.