Tripple integrals, converting into spherical coordinates issue~

In summary, the problem involves evaluating a triple integral over a region B in spherical coordinates. The solution involves converting from x, y, z coordinates to u, v, w coordinates and using the Jacobian determinant to account for the difference in change. The extra factor of p^2*sin(phi) in the formula is needed for this conversion. Additional resources and tutorials can be found online for further understanding.
  • #1
mr_coffee
1,629
1
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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  • #2
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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  • #3
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
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.
 

1. What are triple integrals and how are they different from regular integrals?

Triple integrals are a type of integral that involves integrating over three variables, usually denoted as x, y, and z. It is different from regular integrals, which only involve one variable, because it involves finding the volume under a three-dimensional surface instead of just the area under a two-dimensional curve.

2. Why is converting triple integrals into spherical coordinates useful?

Converting triple integrals into spherical coordinates can be useful because it simplifies the integral by eliminating one of the variables. This can make it easier to solve and can also help to visualize the problem in a more intuitive way.

3. How do you convert a triple integral into spherical coordinates?

To convert a triple integral into spherical coordinates, you need to use the following equations:
x = ρsinφcosθ
y = ρsinφsinθ
z = ρcosφ
where ρ is the distance from the origin, φ is the angle from the positive z-axis, and θ is the angle from the positive x-axis. Then, you can replace the x, y, and z values in the integral with these equations and integrate over the new variables ρ, φ, and θ.

4. What are the limits of integration when converting into spherical coordinates?

The limits of integration when converting into spherical coordinates depend on the shape of the region being integrated. Typically, the limits for ρ will be from 0 to some value r, the limits for φ will be from 0 to π, and the limits for θ will be from 0 to 2π. However, these limits may vary depending on the orientation and size of the region.

5. How do you know when to use spherical coordinates for a triple integral?

You should use spherical coordinates for a triple integral when the region being integrated has spherical symmetry or when the problem is easier to visualize and solve using spherical coordinates. It is also useful when one of the variables in the original integral can be eliminated by using the equations for spherical coordinates.

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