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please help i only know how to do around $x=0$ and $y=0$.

$A = \{ (x,y); 1/2 \leq x \leq 2 , 0 \leq y \leq 1/x^2\}$

Find the volume of the solid obtained by rotating around $y=-1$ and $x=3$

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up vote 2 down vote accepted

I have to draw a picture to solve such a problem.

Do the first volume by taking slices perpendicular to the $x$-axis.

So take a slice "at" $x$, of width "$dx$."

The cross-section is a circle. We need the radius of that circle.

By looking at the picture, you can see that the radius is $1+\dfrac{1}{x^2}$. So the area of the circle of cross-section is $\pi\left(1+\frac{1}{x^2}\right)^2$. For the volume of the thin slice, multiply by $dx$. For the whole volume, "add up" (integrate) from $x=1/2$ to $x=2$. Our volume is $$\int_{1/2}^2 \pi\left(1+\frac{1}{x^2}\right)^2\,dx.$$

Another way: So you can handle about the $x$-axis? OK, we will cheat. Draw the picture. Now lift up the function by $1$, getting $y=1+\dfrac{1}{x^2}$. What used to be the nasty line $y=-1$ is the nice $x$-axis. So rotate $y=1+\dfrac{1}{x^2}$ about the $x$-axis!

For the second problem, there are several ways, depending on whether you want to do volume by slicing or by cylindrical shells.

It is easiest by shells. Take a thin vertical strip of the region we are rotating, at $x$, of width $dx$. Now rotate this strip about $x=3$.

We get a cylindrical shell. The radius of the shell is $3-x$. So the circumference is $2\pi(3-x)$. Multiply by the height $\dfrac{1}{x^2}$ and by the thickness $dx$, and "add up." We get $$\int_{1/2}^2 2\pi(3-x)\frac{1}{x^2}\,dx.$$

If you wish to use slicing, you will need to slice perpendicular to the $y$-axis, and integrate with respect to $y$. Not too bad.

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Hint: Since you already know how to do it for $x=0$ and $y=0$, try making a transformation of the function to where it is equivalent to rotating around $x=0$ and $y=0$. To state this more precisely, consider the alternative function $f(x)=\dfrac{1}{(x+3)^2}+1$, where $-\dfrac{5}{2} \leq x \leq -1$, and its volume of revolution around $y=0$ and $x=0$. Is this equivalent to your original question?

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