Surface Integral over a Cone - Stokes?

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Homework Help Overview

The discussion revolves around the application of Stokes' Theorem in the context of evaluating a surface integral over a cone. Participants are exploring the relationship between the theorem and the specific integral setup, questioning the appropriate variables and transformations needed for the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply Stokes' Theorem but express uncertainty about its relevance and the necessary steps. There are discussions about determining the area element dA in terms of dx and dy, and the potential use of polar coordinates to simplify the integral. Questions arise regarding the integrand and the transformation from Cartesian to polar coordinates.

Discussion Status

The discussion is ongoing, with participants providing guidance on the direction of the problem. Some suggest changing to polar coordinates as a viable approach, while others are clarifying the integrand and its relation to the variables involved. There is no explicit consensus on the correct method yet, but several productive lines of reasoning are being explored.

Contextual Notes

Participants are navigating the complexities of the integral setup, including the definitions of functions and the transformations necessary for evaluation. There is mention of specific values and integrals that may not align with expected results, indicating potential misunderstandings or misapplications of the formulas involved.

YayMathYay
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Homework Statement



zuKnP.png


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.
 
Last edited:
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YayMathYay said:

Homework Statement



zuKnP.png


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.
 
Last edited:
Dick said:
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:

eq0002M.gif


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.
 
YayMathYay said:
I also am trying this formula:

eq0002M.gif


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?
 
Dick said:
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.
 
YayMathYay said:
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?
 
Dick said:
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.
 
YayMathYay said:
Yes. But if I change to polar, I get 18pi, because of the 2 there.

Which 2? Show your work, ok?
 
Dick said:
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]
 
  • #10
YayMathYay said:
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.
 

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