Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. (weird equation) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.
$$y = e^{-x^2}, y = 0, x = -1, x = 1$$
(a) about the $x$-axis (b) about $y = -1$
I could normally do one of these problems except the equation is a substitution equation, and this is a rotation problem.. I've never seen anything like it before.. I know how to do substitution and find the integral but this is a rotation problem... how do I go about doing this type of strange problem?
 A: I will assume that you know how to set up integrals for solid of revolution calculations. 
Draw a picture. For the first volume, if we take a cross-section perpendicular to the $x$-axis "at" $x$, the radius of cross-section is $e^{-x^2}$. So our volume is 
$$\int_{-1}^1 \pi(e^{-x^2})^2\,dx.$$
I would rather use the symmetry, and say the volume is twice the volume when we rotate the part from $x=0$ to $x=1$. So we get that our volume is 
$$2\pi \int_0^1 e^{-2x^2}\,dx.$$
Now comes the problem you may have bumped into: you were not able to find an antiderivative of $e^{-2x^2}$. Well, in terms of elementary functions, neither can I. 
So you will have to evaluate the integral numerically. It may be that you have a fancy calculator that does that sort of thing. I you don't, you will have to use a numerical method, like Simpson's Rule. Or else you will use power series to get a good approximation. 
For the second problem, setting up the integral is much the same, except that the cross-section is a "washer," that is, a disk with a hole in it.
The outer radius of a typical cross-section is $e^{-x^2}-(-1)$. There is a hole of radius $1$. 
So the cross-sectional area is $\pi((e^{-x^2}+1)^2-1^2)$. The volume is
$$\int_{-1}^{1}\pi ((e^{-x^2}+1)^2-1^2)\,dx.$$
Again we are in a situation where we cannot find an elementary antiderivative. 
Remark: Oddly enough, if we are rotating about the $y$-axis, we do get an elementary antiderivative. 
