Q:

Word integral problem check. Did I do this correctly?

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

Here are the constraints:
$y = e^x, y = 0, x = -1, x = 1$; about the $x-axis$
Here is the integral I setup. I think the integral involves integrating a solid from x = -1 to x = 1. Right?:
$$\int_{-1}^1 \pi(e^x)^2$$
so $ du = e^x$ and so $\frac{du}{dx} = e^x$ and $du = e^x\cdot dx$
so 
$$\pi \int_{-1}^{1} u \cdot du$$
$$\pi (\frac{u^2}{2} + \frac{u^2}{2})$$
$$=\pi u^2$$
$$= \pi(e^x)^2$$

A:

The initial set up is correct, by the change of variable we have $u=e^x\implies du=e^x dx$ and then
$$\int_{-1}^1 \pi(e^x)^2\, dx=\int_{-1}^1 \pi e^x e^x\, dx=\int_{1/e}^e \pi u\, du=\pi[u^2/2]_{1/e}^e=\frac \pi 2\left(e^2 -\frac1 {e^2}\right)$$

