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

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

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 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.
 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?