# Stokes' Theorem problem

• mathman44
In summary: I took the curl on the cirlce in the xy plane, so z=0.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...

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

mathman44 said:

## 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.

Is the answer you got 0?

mmmboh said:
Is the answer you got 0?

Me? Yes, I got 0.

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?

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 ?

Dustinsfl said:
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.

mathman44 said:
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.

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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mathman44 said:
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.

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

Shameless bump

mathman44 said:
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.

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

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...

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

I did my integral a different way.

Which way?

mmmboh said:
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.

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.

ideasrule said:
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.

mmmboh said:
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] vela said: 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. 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 $$8\pi$$ ideasrule said: 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?

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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gabbagabbahey said:
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.

## 1. What is Stokes' Theorem?

Stokes' Theorem is a mathematical theorem that relates the integral of a vector field over a surface to the line integral of the same vector field around the boundary of that surface.

## 2. How is Stokes' Theorem used in real-world applications?

Stokes' Theorem is commonly used in physics and engineering to calculate flux, circulation, and other related quantities in vector fields.

## 3. What is the difference between Stokes' Theorem and Green's Theorem?

Stokes' Theorem is a generalization of Green's Theorem, which only applies to two-dimensional surfaces in the plane. Stokes' Theorem can be applied to surfaces in any dimension.

## 4. What are the prerequisites for understanding and solving Stokes' Theorem problems?

A solid understanding of vector calculus, including vector fields, line integrals, and surface integrals, is necessary to understand and solve Stokes' Theorem problems.

## 5. What are some common mistakes made when applying Stokes' Theorem?

Some common mistakes include not correctly identifying the surface and its boundary, not correctly setting up the line and surface integrals, and not taking into account the orientation of the surface. It is important to carefully follow the steps of the theorem to avoid these mistakes.

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