Converting triple integral coordinates

[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 coordinatesI 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?

 

[LCKurtz]

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

 

[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

 

[LCKurtz]

Yes, you've got it.

 

[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

 

[LCKurtz]

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