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Find the volume of the solid that results when the regions bounded by $y=x^3, x=2$, and the x-axis is revolved around the line $x=2$?

I don't really understand what it meant by "x-axis is revolved around the line $x=2$"

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Bounded by $y=x^3$, $x=2$, and the $x$-axis. Dot. Then revolve around the line $x=2$. – 1015 Mar 20 '13 at 22:21
The correct answer I see is 16pi/15. But I'm no where close to finding it. – user67740 Mar 20 '13 at 22:41
Have you been taught the disk method? The shell method? If not, I understand why you are nowhere close to finding the answer. – 1015 Mar 20 '13 at 22:47
Could you show me the steps to the answer? oh and I misread the answer, it should be 16pi/5. Also I solved for Y value instead of X, but I got a answer of 20pi. – user67740 Mar 20 '13 at 23:09
up vote 2 down vote accepted

It is [the region bounded by $y=x^3$, $x=2$, and the $x$-axis] is [rotated about the line $x=2$]. (Admittedly, the wording is not optimal, but from the context the intended meaning can be deduced.)

In an exam, where one has to be particularly careful about possible ambiguities, one might use the following wording. Let $\mathcal{A}$ be the finite region bounded by the curve $y=x^3$, the line $x=2$, and the $x$-axis. Find the volume of the solid of revolution obtained when $\mathcal{A}$ is rotated about $x=2$.

Remark: To find the volume, first draw a picture of the region.

We can then use slices perpendicular to the $y$-axis, and end up integrating with respect to $y$.

But it is more pleasant to use the method of cylindrical shells.

Added: We do the problem by slicing, which is the harder way. Take a slice perpendicular to the $y$-axis, at height $y$. The cross-section is a circle, with radius $2-x$, that is, $2-y^{1/3}$. Note that $y$ travels from $0$ to $8$. So our volume is $$\int_0^8 \pi(2-y^{1/3})^2\,dy.$$ To calculate, expand $(2-y^{1/3})^2$ as $4-4y^{1/3}+y^{2/3}$, and integrate term by term.

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I integrated the equation with respect to Y. But I got a answer of 20pi. Can you show me steps so I can see what I did wrong – user67740 Mar 20 '13 at 23:20
So I see, I thought I need to square them separately, which I got it wrong as $(2)^2$ - $(Y^$1/3$)^2$ – user67740 Mar 20 '13 at 23:41
I set up the "slicing" (cross-sections) integral. If you set up the same integral, there was just a minor slip in execution. If you set up a different integral, your radius is probably wrong. – André Nicolas Mar 20 '13 at 23:42
I get it now. Thank you ! – user67740 Mar 20 '13 at 23:42

There will be a region between the lines $y=x^3$, $x=2$ and $y=0$ (the last one is the $x$ axis). If you revolved that region around the line $x = 2$, you would have a 3D region. You are being asked to find the volume of that region.

The region looks like the almost-triangle in this chart. you're going to revolve it around the line $x=2$.

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