# Tripple integrals, converting into spherical coordinates issue~

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,
http://tutorial.math.lamar.edu/AllBrowsers/2415/SphericalCoords_files/eq0020M.gif [Broken]

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: http://tutorial.math.lamar.edu/AllBrowsers/2415/SphericalCoords_files/eq0022M.gif [Broken]
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!

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## Answers and Replies

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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.

HallsofIvy
Science Advisor
Homework Helper
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.