Surface Integral over a Cone - Stokes?

In summary: And when you integrate it over r, you get 6r dr dt instead of 2(x^2+y^2).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. And when you integrate it over r, you get 6r dr dt instead of 2(x^2+y^2).
  • #1
YayMathYay
22
0

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 [itex]\grad F[/itex] to take? I know this is related to Stoke's (I hope), but not sure how to begin.
 
Last edited:
Physics news on Phys.org
  • #2
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 [itex]\grad F[/itex] 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:
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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.
 
  • #8
YayMathYay said:
Yes. But if I change to polar, I get 18pi, because of the 2 there.

Which 2? Show your work, ok?
 
  • #9
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.
 

1. What is a surface integral over a cone?

A surface integral over a cone is a mathematical operation that calculates the total flux or flow of a vector field through the surface of a cone. It is also known as a flux integral or a cone integral.

2. How is a surface integral over a cone different from a regular surface integral?

A surface integral over a cone is different from a regular surface integral in that the surface being integrated over is a cone instead of a flat surface. This means that the surface must be parametrized differently and the limits of integration must also be adjusted accordingly.

3. What is the significance of a surface integral over a cone in physics and engineering?

A surface integral over a cone has many applications in physics and engineering, particularly in the study of fluid flow and electromagnetism. It is used to calculate the flow of a fluid through a cone-shaped object, such as a cone-shaped pipe or nozzle, and to calculate the electric or magnetic flux through a cone-shaped surface.

4. How is a surface integral over a cone related to Stokes' theorem?

Stokes' theorem is a mathematical theorem that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field along the boundary of the surface. A surface integral over a cone is a special case of Stokes' theorem, where the surface being integrated over is a cone-shaped surface.

5. What are some common techniques for evaluating a surface integral over a cone?

Some common techniques for evaluating a surface integral over a cone include using parametrization to express the surface in terms of two variables, using symmetry to simplify the integral, and using the divergence theorem to convert the surface integral into a volume integral. Additionally, numerical methods such as Monte Carlo integration can also be used for more complex surfaces.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
987
  • Calculus and Beyond Homework Help
Replies
4
Views
964
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
968
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top