Calculating Volume of Solid Bounded by Cylinders and Plane

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SUMMARY

The volume of the solid bounded above by the cylinder defined by the equation z = 4 - x², on the sides by the cylinder x² + y² = 4, and below by the xy-plane is calculated using the integral ∫_{-2}^{2}∫_{-√(4 - x²)}^{√(4 - x²)}∫_{0}^{4 - x²} dz dy dx = 12π. This integral setup is confirmed to be correct. For finding the average value of the function f(x, y, z) = √(xyz) within the same solid, the integral is set up as (1/12π)∫_{-2}^{2}∫_{-√(4 - x²)}^{√(4 - x²)}∫_{0}^{4 - x²} √(xyz) dz dy dx, which is also validated as correct.

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VinnyCee
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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 :smile:
 
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Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.
 
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?
 

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