Thread: am i doing this right? triple integral, spherical coordinates

so what i am concerned about here is simply setting up the correct bounds. if i do that right, i know how to integrate just fine.
i need to find the bounds for the hemisphere given by x^2 + y^2 + z^2 = 16 and z>=o
solving for z, i get z= root(6-x^2-y^2. )

i believe that is a sphere, right?

so a sphere that lies above (or with the bottom directly on xy plane) with radius of 4.

so my bounds for 0<roe<4

theta is the angle on the xy plane, and since it's a sphere, it goes all the way around, from 0 to 2pi.

and for phi, that goes from the z axis down to the y axis. and since it can't go below the y axis into the negative z, (because z>=0), that would mean the max is pi/2.

so is that correct for my bounds?

0<roe<4
0<theta<2pi
0<phi<pi/2

i don't quite understand this, but i'm trying to picture this 3 dimensionally. anyway, if i am wrong, could someone point out where i am going wrong? thank you

Everything looks good to me, except that you should spell it rho, not roe.

ha good call. thanks for the response!