How would I perform this surface integral?

In summary, the homework equation is F = r^3 r i converted the cartesian coordinates to spherical in F.
  • #1
Boltzman Oscillation
233
26

Homework Statement


∫∫ Fndτ over the spherical region x^2 + y^2 + z^2 = 25
given F = r^3 r i already converted the cartesian coordinates to spherical in F

Homework Equations


n = r[/B]

The Attempt at a Solution



I know I can plug in F into the equation and then dot it with r to get the following:

∫∫ r^3 dτ
r should be constant at r = 5 so i can take it out of the integral and substitute
5^3∫∫dτ
now how can i solve the ∫∫dτ? I would have to convert dτ into spherical coordinates? Note: the double integral should have a circle on it but I don't know how to add one here. Any help is useful.
 
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  • #2
##d\tau## in this case is the area element for the surface of the sphere. It should be in spherical coordinates, yes. If you don't know what it is, you can calculate the volume element in spherical coordinates from the Jacobian, then make ##r## constant (thereby removing ##dr##).
 
  • #3
Dewgale said:
##d\tau## in this case is the area element for the surface of the sphere. It should be in spherical coordinates, yes. If you don't know what it is, you can calculate the volume element in spherical coordinates from the Jacobian, then make ##r## constant (thereby removing ##dr##).
Ah so dτ = r^2sinΦdθdΦ, plugging in r = 5 and then into the equation I get:

5^5∫∫sinΦdθdΦ

solving the integrals i get
5^5(θ)cos(Φ)
 
  • #4
Close. But if those are closed integrals, ##\oint## (pretend that's a double), it isn't an indefinite integral.
 
  • #5
Boltzman Oscillation said:
how can i solve the ∫∫dτ?
Isn't it just the surface area of the sphere? Or do I misunderstand the notation?
 
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  • #6
ah so i should consider my boundaries for a sphere? Where θ is from 0 to 2pi and φ is from 0 to pi.

answer should be 6250pi
 
  • #7
You're off by a factor of 2, but yes, the method is correct.
 
  • #8
Got it, Thanks Dewgale.
 
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Related to How would I perform this surface integral?

1. What is a surface integral?

A surface integral is a type of integral in multivariable calculus that is used to calculate the flux of a vector field across a surface. It involves integrating a function over a two-dimensional surface.

2. How do I determine the limits of integration for a surface integral?

The limits of integration for a surface integral depend on the type of surface being integrated over. Typically, the surface is described using a parametric equation and the limits are determined by the range of the parameters.

3. What is the difference between a single and double surface integral?

A single surface integral is used to integrate over a two-dimensional surface, while a double surface integral is used to integrate over a three-dimensional surface. The limits of integration and the integrand are different in each case.

4. How do I choose the appropriate surface element for a surface integral?

The surface element for a surface integral is chosen based on the type of surface being integrated over. For example, a flat surface would use a rectangular element, while a curved surface would use a curvilinear element.

5. Can a surface integral be solved using any integration technique?

No, a surface integral cannot be solved using any integration technique. It requires specific techniques such as parameterization, substitution, or using surface elements. The appropriate technique depends on the complexity of the surface and the integrand.

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