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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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    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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Hint: You can define $z=y-3$ and rewrite everything in $z$, then rotate around $z=0$. Alternately you can review how the cylindrical shell integration formula is derived and modify it for rotation around $y=3$. A bit of the area from $y$ to $y+\Delta y$ that extends from $x_1$ to $x_2$ contributes $(x_2-x_1) 2 \pi (y-3) \Delta y$ to the volume.

Added: I would also use the washers, but the title asked for cylinders. If you rotate around $y=3$, you are rotating around a line parallel to the $x$ axis, so your washer is formed by a rectangle $\Delta x$ by $y_1$ to $y_2$. The inner and outer radii of the washer have to be measured from $y=3$, so the integral is $\int_{-2}^0 \pi[(3-y_1)^2-(3-y_2)^2]dx$. P. S. I don't see an effect of $y=0$ on the region, it is bordered by $y=x^3, x=0, y=-8$.

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    Yes, I switched the bounds. $y=-8$ is further from $y=3$ and should be the positive term. Altogether I get $\int_{-2}^0 \pi[(3-y_1)^2-(3-y_2)^2]dx=\int_{-2}^0 \pi(11^2-(3-x^3)^2)dx$, and http://www.wolframalpha.com/input/?i=integrate+pi%28121-%283-x^3%29^2%29+from+-2+to+0 gets $1272 \pi/7$2011-12-07