# Triple integral in spherical coordinates

## 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.

fzero
Homework Helper
Gold Member
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

$$z = 1 + \rho \cos\theta.$$

The ranges of $$(\rho,\theta,\phi)$$ don't change, but $$\rho^2 = x^2 + y^2 + (z-1)^2$$

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.

Last edited:
fzero
Homework Helper
Gold Member
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, $$\rho = 0$$ is the point (0,0,0) in Cartesian space. If you use a spherical coordinate system where we shift

$$z= 1 + \rho \cos(\phi),$$

then $$\rho = 0$$ 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.

If you didn't shift it, how would you find the bounds of rho?

fzero
Homework Helper
Gold Member
If you didn't shift it, how would you find the bounds of rho?

In the regular coordinates where you have

$$\rho = \sqrt{x^2+y^2+z^2} ,$$

you see that, because $$-\infty<x,y,x<\infty$$, we have $$0<\rho < \infty.$$ In the shifted coordinates, we find the same range of values for $$\rho$$.

HallsofIvy
$\rho$ 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 $\rho$ while the second has distance 2 from (0, 0, 0) and is the largest value of $\rho$.