How to Use Stokes' Theorem to Solve a Sphere Integration 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]kAttempt 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

 

[mathman44]

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

 

[ideasrule]

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?

 

[ideasrule]

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 ?

 

[ideasrule]

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.

 

[ideasrule]

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'...

 

[gabbagabbahey]

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

 

[gabbagabbahey]

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.

 

[vela]

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...

 

[ideasrule]

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

 

[mmmboh]

I did my integral a different way.

 

[ideasrule]

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.

 

[vela]

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.

 

[vela]

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

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

 

[gabbagabbahey]

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]

 

[gabbagabbahey]

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

 

[gabbagabbahey]

Are we all integrating <br /> \int -4(sin\theta)^2 + (2cos\theta)^4 <br /> 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 n...Perhaps this is the source of your error?

 

[ideasrule]

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.

 

[ideasrule]

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.