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 September 13th, 2017, 04:38 PM #1 Member   Joined: Feb 2017 From: East U.S. Posts: 40 Thanks: 0 Find the volume of the solid whose base is the region... Find the volume of the solid whose base is the region in the x-y plane bounded by y=x^2, y=9, and x=0, and whose cross sections perpendicular to the x axis are: a) squares b) rectangles with a height of 4 I missed a single class and this is on the homework which I have no idea how to do. I've never even heard of a cross section so I assume this is something they talked about in class. Can someone at least show me how to set up the problem and I think I can figure out the answer from there. Thanks, everyone.
 September 13th, 2017, 06:40 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 2,982 Thanks: 1575 Hopefully, you've sketched the base of the described solids that lie in the x-y plane. The side of the square cross-section for the first solid has length $(9-x^2)$ ... $\displaystyle V = 2\int_0^3 (9-x^2)^2 \, dx$ The second described solid has a rectangular cross-section with base $(9-x^2)$ and height $4$ ... you set this one up.
September 13th, 2017, 08:23 PM   #3
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 Originally Posted by skeeter Hopefully, you've sketched the base of the described solids that lie in the x-y plane. The side of the square cross-section for the first solid has length $(9-x^2)$ ... $\displaystyle V = 2\int_0^3 (9-x^2)^2 \, dx$ The second described solid has a rectangular cross-section with base $(9-x^2)$ and height $4$ ... you set this one up.
Ok, so the second one would be:

$\displaystyle V = 2\int_0^3 4(9-x^2) \, dx$

September 14th, 2017, 03:23 AM   #4
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 Originally Posted by nbg273 Ok, so the second one would be: $\displaystyle V = 2\int_0^3 4(9-x^2) \, dx$
Correct ... I hope you noted the use of each solid's symmetry w/respect to the y axis in setting the limits of integration.

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