Homework Help: Converting triple integral coordinates

1. Dec 13, 2009

619313

1. consider the triple integral (x^2 +Y^2) dV where it is bounded by a solid sphere of radius R. Set up the integral using rectangular coordinates

I tried setting this up with the bounds [ -sqrt(R^2-x^2-Y^2) <= Z <= sqrt(R^2-x^2-Y^2) ,
-R <= X <= R , -sqrt(R^2-x^2) <= Y <= sqrt(R^2-x^2) ] am I on the right path???

Last edited: Dec 13, 2009
2. Dec 13, 2009

LCKurtz

Yes, but make sure you have them in the correct order when you write down the integrals.

3. Dec 13, 2009

619313

Fantastic! Yea I know that it goes dz dy dx
now How about if I rewrite to cylindrical coordinates?
0<= theta <= 2pi , 0<=r<=R , -sqrt(R^2-r^2)<=z<= sqrt(R^2-r^2) with the integrand being r^3 dz dr dtheta

4. Dec 13, 2009

LCKurtz

Yes, you've got it.

5. Dec 13, 2009

619313

first, thanks so much for the help

just to make sure I fully understand all of these triple integrals, to set it up in spherical I should get
0<= theta<= 2pi, 0<= phi <= pi, 0<= p <= R,
where the integrand is (p^4) (sin(FI))^3 dp dphI dtheta

6. Dec 13, 2009

LCKurtz

Yes. Very nice. And welcome to the Physics forum.