Surface integral parametrization

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SUMMARY

The discussion focuses on evaluating the surface integral \(\iint_S y \; dS\) over the portion of the sphere defined by \(x^2 + y^2 + z^2 = 1\) that lies above the cone \(z = \sqrt{x^2 + y^2}\). Participants clarify that the parametrization of the sphere in spherical coordinates is given by \(r = \langle \cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi \rangle\), where \(\phi\) is the angle from the positive z-axis. The cone's role is limited to determining the integration limits for \(\phi\), specifically that \(\phi\) must be \(\pi/4\) for points on the cone, but does not need to be included in the parametrization of the surface itself.

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  • Spherical coordinates in calculus
  • Understanding of surface integrals
  • Knowledge of limits of integration in multivariable calculus
  • Familiarity with the geometric interpretation of cones and spheres
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flyingpig
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Homework Statement



Evaluate the surface integral

\iint_S y \; dS

S is the part of the sphere x^2 + y^2 + z^2 = 1 that lies above the cone z=\sqrt{x^2 + y^2}

The Attempt at a Solution



I know to use spherical coord so I did

r = <\rho cos\theta sin\phi, \rho sin\theta sin\phi, ?>

The book did r = < cos\theta sin\phi, sin\theta sin\phi, cos\phi>

I don't understand... why does z = \rho cos\phi?
 
Last edited:
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No I know what it means but this is parametrization is just the sphere isn't it? It doesn't include the cone
 
*Sorry I meant without the rho
 
flyingpig said:
*Sorry I meant without the rho

rho=1 if the sphere is x^2+y^2+z^2=1, isn't it?
 
Yes, but the z component still confuses me, they didn't include the cone
 
flyingpig said:
Yes, but the z component still confuses me, they didn't include the cone

The cone tells you about a limit on phi. What's phi for any point on the cone? What are the phi values for points on the sphere above the cone?
 
pi/4, but I am still confuse about the parametrization coudl you run it through like you are doing the problem?
 
flyingpig said:
pi/4, but I am still confuse about the parametrization coudl you run it through like you are doing the problem?

You should know that's not the way it works by now. You should be able to figure out the limits on phi and theta now. So all that's left is to figure out what y and dS are on the sphere in terms of phi and theta and do the integration.
 
  • #10
No, my problem is with parametrizing the surface. I stil ldon't understand the z-component of the parametric surface
 
  • #11
flyingpig said:
No, my problem is with parametrizing the surface. I stil ldon't understand the z-component of the parametric surface

You are parameterizing the surface in terms of phi and theta. phi and theta determine x, y and z. What's so mysterious about the z-component? It's cos(phi). Do you understand the x-component?? To do the integral you only need dS and y. You don't even need z.
 
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  • #12
flyingpig said:

Homework Statement



Evaluate the surface integral

\iint_S y \; dS

S is the part of the sphere x^2 + y^2 + z^2 = 1 that lies above the cone z=\sqrt{x^2 + y^2}

flyingpig said:
No I know what it means but this is parametrization is just the sphere isn't it? It doesn't include the cone

flyingpig said:
Yes, but the z component still confuses me, they didn't include the cone

Your parametrization does not need to include the cone because the problem concerns the surface S, which is just part of the sphere. As Dick was saying, the cone is only relevant to your limits of integration- you do not need to parametrize the cone because it is not part of S.
 

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