Solid of revolution volume, washer method.

lmerry

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Nov 2, 2006
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Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y=4.

y=x\displaystyle y=x
y=3\displaystyle y=3
x=0\displaystyle x=0

so, here's what i've done. I know the answer i got is wrong, but i can't see where i slipped up. However, this is new stuff to me so im not too comfortable with it yet.

V=π([R(x)]2[r(x)]2)dx\displaystyle V=\pi \int([R(x)]^2-[r(x)]^2)dx

outer radius R(x) i think should be 3-x

inner radius r(x) seems to be a constant 1

so therefore, it seems to me that the following would give me the correct answer.

V=π([3x]2[1]2)dx\displaystyle V=\pi \int([3-x]^2-[1]^2)dx (from 0 to 3)

(I dont know the tex coding very well so i dont know how to make it display a definite integral with bounds.)

this gives the answer 6π\displaystyle 6\pi but my text says it's supposed to be 18π\displaystyle 18\pi

so where did i mess up?

thanks.
 
Must you use washers?. How about shells?.

\(\displaystyle \L\\2{\pi}\int_{0}^{3}y(4-y)dy\)


But, if you must use washers:

\(\displaystyle \L\\{\pi}\int_{0}^{3}[(x-4)^{2}-1]dx\)


Here's a graph of the washers. See them in the animation?.

rotate5rq9.gif
 
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