# Volume bounded above by cylinder z = 4 - x^2, x^2 + y^2 = 4 on sides,xy-plane bottom

1. Apr 13, 2005

### VinnyCee

Here is the problem:

Find the volume of the solid that is bounded above by the cylinder $$z = 4 - x^2$$, on the sides by the cylinder $$x^2 + y^2 = 4$$, and below by the xy-plane.

Here is what I have:

$$\int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;dz\;dy\;dx\;=\;12\pi$$

Is that correct? I didn't post the many steps for integration, but the integral calulation is correct, I just need to know if I set up the integral right. Thanks again :rofl:

2. Apr 13, 2005

### HallsofIvy

Staff Emeritus
Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.

3. Apr 13, 2005

### VinnyCee

What if f(x, y, z) = sqrt(xyz), how to find average value?

Thanks for the double checking! The next problem uses this same integral and assumes that $$f\left(x, y, z\right) = \sqrt{x\;y\;z}$$. Then it says to setup the integral to find the average value of the function within that solid.

Here is what I have:

$$\frac{1}{12\pi}\;\int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;\sqrt{x\;y\;z}\;dz\;dy\;dx$$

Does that look right?