# Triple Integral Conversion (Spherical, Rectangular, Cylindrical)

• Nov 20th 2007, 04:42 PM
numberonenacho
Triple Integral Conversion (Spherical, Rectangular, Cylindrical)
http://img101.imageshack.us/img101/6167/69688576jh3.gif

So I am supposed to find an equivalent integral in spherical coordinates,
rectangular coordinates and then evaluate the integral.

I know im supposed to use the rules z = rho cos phi
r = rho sin phi

for spherical, but do I just substitute it in? And what do I change the range of the integrals to.

And for rectangular I know I use x = r cos theta
y = r sin theta
z = z
But also I am not suer what to change the ranges to.
And then from there I'll have to evaluate the integral.

I have also made an identical thread in the calculus section. I hope Mods don't mind.
• Nov 20th 2007, 04:54 PM
ThePerfectHacker
$\displaystyle \int_{\pi/2}^{\pi} \int_0^2 \int_0^{\sqrt{4-r^2}}r^2 \cos \theta dz~dr~d\theta = \iint_R \int_0^{4-x^2-y^2} x dz~dA$ where $\displaystyle R$ is the disk with radius $\displaystyle 2$ from $\displaystyle \theta = \pi/2$ to $\displaystyle \theta = \pi$.
And this is equivalent to,
$\displaystyle \iiint_V xdV$
Where $\displaystyle V$ is the volume made of the sphere with radius $\displaystyle 2$ bounded by the $\displaystyle xy$ plane (the upper half) and making angles $\displaystyle \theta = \pi/2$ to $\displaystyle \theta = \pi$.
In spherical coordinates that is,
$\displaystyle \int_{\pi/2}^{\pi} \int_0^{\pi/2} \int_0^2 \rho \sin \phi \cos \theta (\rho^2 \sin \phi) d\rho~d\phi~d\theta$
• Nov 20th 2007, 07:21 PM
numberonenacho
• Nov 20th 2007, 08:00 PM
ThePerfectHacker
Quote:

Originally Posted by numberonenacho
So thats the Rectangular coordinates? What would the limits be? 0-2 pi/2 - pi ?
And then I would evaluate that integral? would it matter which one I used, all of them should come out the same answer right?

The last integral is in spherical coordinates. What I did I converted the cyclindrical form into rectangular and then rectangular into spherical, like the problem want you to do. You can use any of the integrals to find the answer it will all be the same.
• Nov 24th 2007, 08:28 AM
numberonenacho
When I evaluated the integral, I got an answer of -8. Does that seem right?
I can post up my work if you would like.