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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
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VinnyCee
#1
Apr13-05, 07:29 PM
P: 492
Here is the problem:

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

Here is what I have:

[tex]\int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;dz\;dy\;dx\;=\;12\pi[/tex]

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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HallsofIvy
#2
Apr13-05, 09:06 PM
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Thanks
PF Gold
P: 39,552
Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.
VinnyCee
#3
Apr13-05, 11:25 PM
P: 492
Thanks for the double checking! The next problem uses this same integral and assumes that [tex]f\left(x, y, z\right) = \sqrt{x\;y\;z}[/tex]. Then it says to setup the integral to find the average value of the function within that solid.

Here is what I have:

[tex]\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[/tex]

Does that look right?


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