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

by VinnyCee
Tags: bounded, cylinder, sides, volume, xyplane
 P: 492 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
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,900 Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.
 P: 492 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?

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