# Homework Help: Triple integral?

1. May 4, 2006

### pivoxa15

I like to use cartesian coords

Find the region to the intersecting cyclinders x^2+y^2<=a^2 and x^2+z^2<=a^2

What I have trouble finding is the domain of integration

Currently I have
a to -a for dx
-srt(a^2-x^2) to srt(a^2-x^2) for dy
-srt(x^2+y^2) to srt(x^2+y^2) for dz

But this integration turned out to be too complicated to evaluate hence probably wrong.

2. May 4, 2006

### HallsofIvy

"Find the region to ..." ? I assume you mean to find the volume of that region.
It's not at all clear to me where you would get "-srt(x^2+y^2) to srt(x^2+y^2) for dz". I don't see any reason to think that z2 is ever equal to x2+ y2. The projection of the figure down on to the xy-plane is the circle x2+ y2= a2 so, yes, the outer integrals would be
$$\int_{x= -a}^a \int_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2- x^2}} dydx$$

But, for every (x,y), z ranges between $-\sqrt{a^2- x^2}$ and $\sqrt{a^2- x^2}$ also. Looks to me like the inner integral should be
$$\int_{z= -\sqrt{a^2-x^2}}^{\sqrt{a^2-z^2}}dz$$
Of course, that could be done simply as the double integral
$$2\int_{x= -a}^a \int_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\sqrt{a^2- x^2}dydx$$

I think it would be even simpler in polar coordinates

3. May 5, 2006

### pivoxa15

I see how you got your solution which yields the correct solution, 16a^3/3.

I eventually used a different method but is slightly wrong with a factor of 'a' too much. I integrated along one cyclinder along one axis and then specified the limits of the other two axis as -a and a. My integral from the inside out is
From -(x^2+z^2) to x^2+z^2 along y
From -a to a along x
From -a to a along z

I got 16a^4/3. Where did I go wrong?