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Not sure exactly how to get started on this:

Find volume of the solid by rotating the region bounded by $y=x^3$, $x=0$, $y=0$ and $y=-8$ around $y=3$.

I know that the graph should be a cubic function, and the shaded portions should be from $y=0$ to $y=-8$ inside the lower portion of the cubic function $y=x^3$, but the rotation around $y=3$ is what is throwing me off. Up until this point it has just been "rotate around y-axis" or "rotate around x-axis".

Does this mean that the height of the cylinder should be $3-x^3$ or $3-y$? If so, this is what I came up with so far, in terms of y (most likely completely wrong, but hey):

$y=x^3 \Rightarrow 3=x^3$ ; $x=3^\frac{1}{3} \Rightarrow x=y^\frac{1}{3}$

So circumference is: $2\pi y^\frac{1}{3}$

Height: $3-y$

Volume: $V=\int_{-8}^0 \left(2\pi y^\frac{1}{3}\right)\left(3-y\right)$

So far so good?

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    There’s absolutely no reason to use shells on this problem; the natural setup is with washers (annuli).2011-12-06
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    I am not sure how to determine when to use one over the other. I am new to this. Thanks.2011-12-06
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    Also, if I do the washer method, I am still not sure how to account for rotating around $y=3$. If it were just a matter of rotating about the y-axis, I think I would end up with $\int_ {-8}^0 \pi \left(y^\frac{1}{3}\right)^2 dy$.2011-12-06

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