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I am trying to find the volume of $ y= x^3$ $y=0$ $x=1$ about $x=2$

I know what the graph looks like, I did that part properly. I am just trying to figure out how to calculate the rest of it. I know that I can find the volume of the $x^3$ part but I do not know who to subtract the other part.

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Could you please clarify the definition of the solid of rotation, is the slice defined as $\{(x,y) \in \mathbb{R}^2 \colon 0 \leqslant y \leqslant x^3, 1 \leqslant x \leqslant 2 \}$. Or did I get it all wrong ? – Sasha Apr 29 '12 at 19:31
I do not know what that notation means. – user138246 Apr 29 '12 at 19:43
I don't understand the definition either. Do you mean rotating the region bounded by $y=x^3$, $y=0$, $x=1$ around the axis $x=2$? – Ayman Hourieh Apr 29 '12 at 19:49 – user138246 Apr 29 '12 at 19:57
up vote 1 down vote accepted

We can use the cylindrical shell method, or the slicing (area of cross-section) method. In this case, shells are easier.

$1$. Take a thin vertical strip, of width "$dx$" with base running from $x$ to $x+dx$. This strip is at distance $2-x$, roughly, from the line $x=2$. When you rotate the strip, you get a "shell" of thickness $dx$, height $x^3$, and perimeter $2\pi(2-x)$. "Add up," $x=0$ to $x=1$. We get volume $$\int_{x=0}^1 2\pi(2-x)x^3\,dx.$$

$2$. Or use slicing, parallel to the $x$-axis, so we will be integrating with respect to $y$. Take a slice of thickness $dy$, at height $y$. The volume of the slice is $dy$ times the area of cross-section. Note the hole. The area of cross-section is $\pi(2-x)^2-\pi$. In terms of $y$ it is $\pi(2-y^{1/3})^2-\pi$. So the volume is $$\int_{y=0}^1 \pi\left((2-y^{1/3})^2-1\right)\,dy.$$ Or else you can forget about the hole for a while, integrate $\pi(2-y^{1/3})^2$, and remove the volume of the hole later. The hole is a cylinder, with volume easy to find without calculus.

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I am not sure how you are getting those numbers in either example, especially the 2-x and the 2pi parts. – user138246 Apr 29 '12 at 20:13
Draw a picture, with the vertical strip with base $x$ to $x+dx$. Rotate it. You get a shell, illustrated in the picture you produced. This shell has thickness $dx$. Since our strip is more or less at $x$, it is at distance $2-x$ from the line $x=2$. The shell thus has radius $2-x$, and therefore circumference $2\pi(2-x)$, since a circle of radius $r$ has circumference $2\pir$. So if you cut the shell vertically, and flatten it, you get something of thickness $dx$, height $x^3$, and length $2\pi(2-x)$. – André Nicolas Apr 29 '12 at 20:20
the 2pi is coming from the circumferance formula? – user138246 Apr 29 '12 at 20:23
Yes, sorry had a LaTeX typo above, wanted $2\pi r$, the circumference formula. The whole thing is I am sure discussed in detail, with good pictures, in your Stewart. In my edition, it is 6.3, Volumes by Cylindrical Shells. – André Nicolas Apr 29 '12 at 20:29
I was trying to find the area of the circle and then subtract the inner circle, how would I do that? – user138246 Apr 29 '12 at 20:50

Slice your region into thin horizontal slices of thickness $\Delta y$. Each slice, when you revolve it around the line $x = 2$, will become a "washer," that is, a thickened disk with a smaller thickened disk removed (see your picture for a typical slice). The thickness of the disks will be $\Delta y$; what will the areas be?

Well, we need to write them as a function of $y$, so first let's write the cubic as $x = y^{1/3}$. Now, the radius of the big disk will be approximately (EDIT: fixed- see below) $2 - y^{1/3}$ if your horizontal slice includes the point $y$. The radius of the little disk is always 1. So the volume is approximately $\sum \pi((2 - y^{1/3})^2 - 1^2)\Delta y$. Taking a limit as the slices get thinner and thinner, you want the integral

$$ \int_0^1 \pi((2-y^{1/3})^2 - 1)dy $$

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Isn't the radius of the big disc $2-y^{1/3}$? – David Mitra Apr 29 '12 at 20:06
yes, you are right! – user29743 Apr 29 '12 at 20:08
fixed - thanks! – user29743 Apr 29 '12 at 20:09

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