Surface Integral over a Cone - Stokes?

[YayMathYay]

Homework Statement

Homework Equations

I'm guessing Stoke's Theorem? However, I'm not sure how to apply it exactly..

The Attempt at a Solution

Looking at Stoke's Theorem, I'm still not sure how to apply it. I'm really just lost as to where to begin; is there even a \grad F to take? I know this is related to Stoke's (I hope), but not sure how to begin.

 

[Dick]

Homework Statement

Homework Equations

I'm guessing Stoke's Theorem? However, I'm not sure how to apply it exactly..

The Attempt at a Solution

Looking at Stoke's Theorem, I'm still not sure how to apply it. I'm really just lost as to where to begin; is there even a \grad F to take? I know this is related to Stoke's (I hope), but not sure how to begin.

It doesn't look like Stoke's to me. I don't see any vectors anywhere. Your first job is to figure out what dA is in terms of dx*dy. Then a change to polar coordinates might make it easier to do the dx*dy integral. It's actually pretty straight forward.

 

[YayMathYay]

It doesn't look like Stoke's to me. I don't see any vectors anywhere. Your first job is to figure out what dA is in terms of dx*dy. Then a change to polar coordinates might make it easier to do the dx*dy integral. It's actually pretty straight forward.

I also am trying this formula:

But I'm not getting to 9pi.. I'm not sure what the difference between f(x, y, z) and f(x, y, g(x, y)) is.

 

[Dick]

I also am trying this formula:

But I'm not getting to 9pi.. I'm not sure what the difference between f(x, y, z) and f(x, y, g(x, y)) is.

That's the right direction. Your g(x,y) is just z. Solve your equation for z^2 for z to get g(x,y). It's sqrt(3)*sqrt(x^2+y^2), right?

 

[YayMathYay]

That's the right direction. Your g(x,y) is just z. Solve your equation for z^2 for z to get g(x,y).

I did, that and my integrand on the right side came out to be 2(x^2 + y^2).

But I don't see how to get to the correct answer from that.

 

[Dick]

I did, that and my integrand on the right side came out to be 2(x^2 + y^2).

But I don't see how to get to the correct answer from that.

Ok. That's good. The easiest way to do it from here is to change to polar coordinates. Did you try that?

 

[YayMathYay]

Ok. That's good. The easiest way to do it from here is to change to polar coordinates. Did you try that?

Yes. But if I change to polar, I get 18pi, because of the 2 there.

 

[Dick]

Yes. But if I change to polar, I get 18pi, because of the 2 there.

Which 2? Show your work, ok?

 

[YayMathYay]

Which 2? Show your work, ok?

Urgh, I just didn't want to type it out because I don't know how to use LaTeX and it looks messy.

Integrand: 2(x^2 + y^2) dx dy

Changed to polar w/:
x = sqrt(3) * cos t
y = sqrt(3) * sin t

New Integrand: 6r dr dt [r from 0 to sqrt(3), t from 0 to 2pi]

 

[Dick]

Urgh, I just didn't want to type it out because I don't know how to use LaTeX and it looks messy.

Integrand: 2(x^2 + y^2) dx dy

Changed to polar w/:
x = sqrt(3) * cos t
y = sqrt(3) * sin t

New Integrand: 6r dr dt [r from 0 to sqrt(3), t from 0 to 2pi]

And there's your problem. The integrand is supposed to be integrated over r. You don't put a specific value of r in. x^2+y^2 is just r^2.