Surface integral parametrization

In summary, the surface integral of y over the part of the sphere x^2 + y^2 + z^2 = 1 above the cone z=\sqrt{x^2 + y^2} is evaluated using a parametrization in spherical coordinates. The parametrization includes a limit on the angle phi, which is determined by the cone, but the cone itself is not included in the parametrization since it is not part of the surface. The z-component of the parametrization is simply cos(phi), while the x and y components are determined by the angles phi and theta. The integral only requires knowledge of dS and y, and does not involve the z component.
  • #1
flyingpig
2,579
1

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]?
 
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  • #3
No I know what it means but this is parametrization is just the sphere isn't it? It doesn't include the cone
 
  • #4
*Sorry I meant without the rho
 
  • #5
flyingpig said:
*Sorry I meant without the rho

rho=1 if the sphere is x^2+y^2+z^2=1, isn't it?
 
  • #6
Yes, but the z component still confuses me, they didn't include the cone
 
  • #7
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?
 
  • #8
pi/4, but I am still confuse about the parametrization coudl you run it through like you are doing the problem?
 
  • #9
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.
 

Related to Surface integral parametrization

1. What is a surface integral parametrization?

A surface integral parametrization is a mathematical method used to calculate the surface area of a three-dimensional surface by breaking it down into smaller, two-dimensional parts. It involves defining a set of parameters that describe the surface, and then integrating over these parameters to find the total surface area.

2. How is a surface integral parametrization used in physics?

In physics, surface integral parametrization is used to calculate the flux of a vector field through a surface. This is important in many applications, such as calculating the electric or magnetic field flux through a closed surface, or calculating the flow rate of a fluid through a surface.

3. Can any surface be parametrized?

Yes, any smooth, continuous surface can be parametrized. This means that the surface can be described by a set of equations or parameters that vary smoothly over the surface. However, some surfaces may require more complex or challenging parametrizations.

4. How does a surface integral parametrization differ from a line integral parametrization?

A surface integral parametrization is used to calculate the surface area of a three-dimensional surface, while a line integral parametrization is used to calculate the length of a curve in two or three dimensions. Additionally, a surface integral parametrization involves integrating over two parameters, while a line integral parametrization involves integrating over one parameter.

5. What are some real-life applications of surface integral parametrization?

Surface integral parametrization has many real-life applications, including calculating the surface area of a curved roof or a piece of land, determining the flux of a fluid through a surface in engineering or environmental science, and calculating the surface area of a three-dimensional object in computer graphics and animation.

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