Thread: Use triple integral to find volume

Hi guys
please I realy need your help with 3 problems
1- Use triple integral to find the volume of the solid in the first octant bounded by the coordinate planes, the plan y=1, and the parabolic cylinder z= 9-x^2.

2-Find the volume of the region bounded above by the paraboloid z=9-x^2-y^2, below by the xy-plane, and that lies outside the cylinder x^2+y^2=1.

3- (a) find the center of mass of a solid of constant density bounded below by the paraboloid z=x^2+y^2 and above by plane z=4.

(b) Find the plane z=c that divides the above solid into two part of equal volume. (This plane dose not pass through the center of mass).

Please friends help me as best as you can.

Can you draw them first? That helps a lot I think. Just use whatever you have available to draw it. If you have Mathematica or some other software that's even better. The plot below is the first one. It's pretty clear what has to be done right:

It's the integral of the paraboloid over that little blue rectangle in the x-y plane. So x goes from 0 to 3, y goes from 0 to 1, and z goes from the blue to the top of the paraboloid. Can you now convert this into a triple integral?

Originally Posted by shawsend

Can you draw them first? That helps a lot I think. Just use whatever you have available to draw it. If you have Mathematica or some other software that's even better. The plot below is the first one. It's pretty clear what has to be done right:

It's the integral of the paraboloid over that little blue rectangle in the x-y plane. So x goes from 0 to 3, y goes from 0 to 1, and z goes from the blue to the top of the paraboloid. Can you now convert this into a triple integral?

First, thank you sir for your replay, but answer for your Qus. I'm still confusing in order of the integral and limit of integral.