Triple integral in spherical coordinates

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Homework Statement


use spherical coordinates to calculate the triple integral of f(x,y,z) over the given region.

f(x,y,z)= sqrt(x^2+y^2+z^2); x^2+y^2+z^2<=2z

The Attempt at a Solution


Once I find the bounds, I can do the integral. But I'm having trouble with the bounds of rho.

This region is a sphere of radius 1 centered at (0,0,1). If this is true, then theta and phi both range from 0 to 2pi.

Does rho range from 0 to 2, or do I need to use the function of the region in there somewhere?

Thank you.
 

Answers and Replies

  • #2
fzero
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You can change variables in f so that you move the center of the sphere to the origin. Equivalently, you can use spherical coordinates centered at (0,0,1). The definition of x and y are the same, but

[tex] z = 1 + \rho \cos\theta. [/tex]

The ranges of [tex] (\rho,\theta,\phi)[/tex] don't change, but [tex]\rho^2 = x^2 + y^2 + (z-1)^2[/tex]
 
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Where did the (z-1)^2 come from? And having rho^2 in terms of x,y, and z doesn't really help me. x=rho sin(theta)cos(phi), y=rho sin(theta)sin(phi), and z=rho cos(phi).

I don't really understand your response. Sorry. Could you clarify?
Thank you.
 
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  • #4
fzero
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Where did the (z-1)^2 come from? And having rho^2 in terms of x,y, and z doesn't really help me. x=rho sin(theta)cos(phi), y=rho sin(theta)sin(phi), and z=rho cos(phi).

I don't really understand your response. Sorry. Could you clarify?
Thank you.

In your coordinates, [tex]\rho = 0[/tex] is the point (0,0,0) in Cartesian space. If you use a spherical coordinate system where we shift

[tex]z= 1 + \rho \cos(\phi),[/tex]

then [tex]\rho = 0[/tex] is the point (0,0,1), which is the center of your sphere. This is convenient because we don't have weird ranges for our integral, but the expression of f(x,y,x) is a bit more complicated.
 
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If you didn't shift it, how would you find the bounds of rho?
 
  • #6
fzero
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If you didn't shift it, how would you find the bounds of rho?

In the regular coordinates where you have

[tex]\rho = \sqrt{x^2+y^2+z^2} , [/tex]

you see that, because [tex]-\infty<x,y,x<\infty[/tex], we have [tex]0<\rho < \infty.[/tex] In the shifted coordinates, we find the same range of values for [tex]\rho[/tex].
 
  • #7
HallsofIvy
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[itex]\rho[/itex] is the distance from the origin to points on that sphere. Since the sphere has center at (0, 0, 1), it contains both (0, 0, 0) (at distance 1 directly below (0, 0, 1) on the z-axis) and (0, 0, 2) (at distance 1 directly above (0, 0, 1). The first point has distance 0 from (0, 0, 0) and is the smallest value of [itex]\rho[/itex] while the second has distance 2 from (0, 0, 0) and is the largest value of [itex]\rho[/itex].
 

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