HELP!!! Can someone walk me through this triple integral volume problem???

Question

The problem states: Find the volume of the solid bounded by the cylinder x=y^2 and the planes z=0 and z=4-x. HELP ME!!!

Answer 1

The "cylinder" is a parabola in the x-y plane, the z=0 plane IS the x-y plane, the plane z=4-x is a plane parallel to the y-axis and intersects the x-z plane at x = 4 and z = 4.

Consider an element of the volume along the z axis. That volume will be the area of the section of the parabola times dx:

dV = A(z)dz

Find the rest here:
http://img86.imageshack.us/img86/5413/volumeintegralfx3.png

Answer 2

In the xy-plane, x = y² is a parabola with vertex at (0,0) and with the x-axis as its axis, opening in the +x direction. Think of the xy-axis as horizontal. Now imagine moving that parabola vertically, parallel to itself. This is the surface that forms part of the boundary of the solid. z=0 is a plane -- in fact, it's the xy-plane. z=4-x is also a plane. Because its equation doesn't involve y, this plane is parallel to the y-axis. It intersects the xy-plane (which, recall, is z=0) in the line x=4. So the "floor" of the solid is the region in the xy-plane bounded by the parabola x = y² and the line x=4, and the projection of this region onto the "tilted" plane z=4-x forms its "ceiling".

So z goes from 0 to 4-x. I would do z-integration first.

For the x and y integrations, it's just as if you were finding the area of the "floor" with a double integral.