# Tripple integrals, converting into spherical coordinates issue~

1. Mar 5, 2007

### mr_coffee

Hello everyone, this is an example out of the book, but i'm confused on how they got the spheircal cordinates.

Here is the problem:
Evaluate tripple integral over B (x^2+y^2+z^2) dV and use spherical coordinates.

Well the answer is the following:
In spherical coordinates B is represented by {(p,theta,phi)| 0 <= p <= 1, 0 <= theta <= 2pi; 0 <= phi <= pi }; Thus ripple integral of B (x^2+y^2+z^2) dV = tripple intgral (p)^2*p^2 sin(phi) dp d(theta) d(phi)

I'm lost on how they got (p)^2*p^2 sin(phi)

I know the following though,

I figured out how they get the new bounds in spherical coordinates.

But when I used the formula, all i got was
(psin(phi)*cos(theta))^2 + (psin(phi)*sin(theta))^2 + ((pcos(phi))^2; not what they got.

I also saw that:
but this still doesn't explain the extra p^2*sin(phi) it does explain the extra p^2 though.
Any help would be great!

Last edited: Mar 5, 2007
2. Mar 5, 2007

3. Mar 5, 2007

### mr_coffee

I was able to follow the directions on converting to polar coordinates which I have no problem in doing but when I read the small part on spherical I really didn't see them show any conversion at all, i'll try to google to find more tutorials.

4. Mar 6, 2007

### HallsofIvy

Staff Emeritus
In general, if you convert from x,y,z coordinates to u(x,y,z), v(x,y,z), w(x,y,z), then dxdydz= J(u,v,w)dudydz where J(u,v,w) is the "Jacobian" determinant:
$$\left|\begin{array}{ccc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}\end{array}\right|$$

In particular, for spherical coordinates, that is
$$\left|\begin{array}{ccc}cos(\theta)sin(\phi) & -\rho sin(\theta)sin(\phi) & \rho cos(\theta)cos(\phi) \\ sin(\theta)sin(\phi) & \rho cos(\theta)sin(\phi) & \rho sin(\theta) cos(\phi) \\ cos(\phi) & 0 & -\rho sin(\phi)\end{array}\right|= \rho^2 sin(\phi)$$

I'm sure that's covered in your text book.