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It says use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis y=x^3 y=8 about the axis x=3 I drew the graph, reflected it about the x=3 line, and drew a cylinder. `I figured that the radius is just r=3-x and the height would just be h=x^3-0 (since the lowest y value is zero), and plugged these into the integral of 2(pi)(r)(h) from 0 to 2. However, I got the wrong answer (correct answer should be 264pi/5).

I have a feeling that my height may be wrong but I'm not sure why.

Thank you!

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    $\begingroup$ I think you left out some information in the problem above. You say the bounds are $y=x^3$ and $y = 8$ but is it also bounded by the $x$ and $y$ axes? $\endgroup$
    – WaveX
    Commented Dec 7, 2016 at 21:08

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Hint:

Assuming the $x$ and $y$ intercepts are bounds in this problem (because you didn't mention it) the two equations intersect at $x = 0$ and $x = 2$, so those are the bounds of integration. By rotating around the line $x = 3$, the radius of the shell is $3-x$ and the height of the shell is $8-x^3$

Using the shell method, the integral would be set up like so: $$2\pi\int^2_0(3 - x)(8 - x^3) dx $$

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