Tripple integrals, converting into spherical coordinates issue~

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Homework Help Overview

The discussion revolves around the conversion of triple integrals into spherical coordinates, specifically evaluating the integral of the function (x^2+y^2+z^2) over a specified region B. Participants express confusion regarding the transformation and the resulting Jacobian factor in spherical coordinates.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how the spherical coordinates lead to the expression (p)^2*p^2*sin(phi) and questions the derivation of the Jacobian factor. Other participants discuss the need for a correction factor when changing variables and mention the Jacobian determinant in the context of spherical coordinates.

Discussion Status

Participants are exploring various aspects of the conversion process, with some providing references to external resources for further clarification. There is an ongoing examination of the Jacobian and its role in the transformation, but no consensus has been reached on the original poster's specific confusion.

Contextual Notes

Some participants note that the textbook may not provide sufficient detail on the conversion process from Cartesian to spherical coordinates, leading to a search for additional tutorials and explanations.

mr_coffee
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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 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
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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The simple explanation is that a small change in x y and z is not exactly the same change in rho psi and theta. There must be some factor to correct for this.

This explains it nicely:

http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node77.html
 
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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.
 
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:
[tex]\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|[/tex]

In particular, for spherical coordinates, that is
[tex]\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)[/tex]

I'm sure that's covered in your textbook.
 

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