Surface integral parametrization

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Homework Help Overview

The discussion revolves around evaluating a surface integral over a specific region of a sphere defined by the equation x² + y² + z² = 1, which is constrained above a cone described by z = √(x² + y²). Participants are exploring the parametrization of the surface and the implications of the cone on the limits of integration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to understand the correct parametrization of the surface, particularly questioning the z-component in spherical coordinates. There is confusion about how the parametrization relates to the cone and the limits it imposes on the angles phi and theta.

Discussion Status

Some participants have provided clarifications regarding the relationship between the parametrization and the surface integral, emphasizing that the cone's role is limited to defining integration limits rather than being part of the parametrization itself. The discussion is ongoing, with participants expressing confusion and seeking further clarification on specific components of the parametrization.

Contextual Notes

Participants note that the parametrization should focus on the surface of the sphere, and there is a recognition that the cone's relevance is primarily in setting the limits for the integration process. There is an acknowledgment of the need to determine the appropriate values for phi and theta based on the geometry of the problem.

flyingpig
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Homework Statement



Evaluate the surface integral

[tex]\iint_S y \; dS[/tex]

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

The Attempt at a Solution



I know to use spherical coord so I did

[tex]r = <\rho cos\theta sin\phi, \rho sin\theta sin\phi, ?>[/tex]

The book did [tex]r = < cos\theta sin\phi, sin\theta sin\phi, cos\phi>[/tex]

I don't understand... why does [tex]z = \rho cos\phi[/tex]?
 
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

[tex]\iint_S y \; dS[/tex]

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

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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