# Volume enclosed by two paraboloids

1. Nov 6, 2008

Hello,

This is a question in which I'm getting the right answer, I just want to make sure what I'm doing its legitimate, as it seems a bit off to me.

The volume enclosed by
$$z=16[x^{2}+y^{2}], z=8-16[x^{2}+y^{2}]$$

I set my triple integral up as follows:
$$\int^{?r}_{?r}\int^{2\pi}_{0}\int^{8-16r^{2}}_{16r^{2}} r dz d\vartheta dr$$

So, my height is the bottom function to the upper, and obviously completing a full circle.
Now here is what I'm unsure about.

I know I need to either do two integrals for r and add them together 0 to 1 and 0 to -1,
or -1 to 1 so that I'm only left with 2/pi (the answer) but what is the logic behind getting the limits for r?

2. Nov 6, 2008

### tiny-tim

I don't understand what your two integrals for r are

r cannot be negative.

You're taking all possible cylindrical slices, so they're from r = … to … ?

(though it might be easier to take horizontal slices, from z = … to … )

3. Nov 6, 2008

### HallsofIvy

Staff Emeritus
That's a good start.
It should be obvious that the origin, r=0, is contained in the volume so that is your lower limit on r. Now, where do the two graphs, z= 16r^2 and z= 8- 16r^2 intersect? That will give the upper limit on the r integral.

4. Nov 6, 2008