I need to find the surface area of $y = (e^x + e^{-x}) / 2$ from x = -1 to x = 1 revolved about the x-axis.
Final integral (w/o symmetry in mind): $\int_{-1}^{1} 2\pi \frac{(e^x + e^{-x})}{2}|\frac{e^x}{2} - \frac{e^{-x}}{2}|dx$
When integrating with my bounds from [-1,1], why do I get zero? In other words, why is it important to notice symmetry in this problem and what causes my integral to become zero? Further, when integrating from -x to x, is it good practice to draw a graph and check for symmetry, preventing an unintentional zero value?
UPDATE
Was just told solution to this problem is wrong. Will update question if needed.
The Integrand is the function
\begin{align} f(x) = 2\pi \frac{(e^x + e^{-x})}{2}\left|\frac{e^x}{2} - \frac{e^{-x}}{2}\right| \end{align}
You'll notice that, if you replace substitute $-x$ for $x$, you get the same function. In other words.
\begin{align} f(x) = f(-x) \end{align}
Intuitively, The height at all $-x$ is the same as the value at $x$. That means that the negative side and the positive side of the graph are mirror images. Thus, to integrate from $-1$ to $1$, you can simply integrate from $0$ to $1$, and double it.
Mathematically,
\begin{align} \int_{-1}^{1} f(x)dx &= \int_{-1}^0 f(x)dx + \int_{0}^1 f(x)dx\\ & = -\int_{1}^{0} f(-x) dx + \int_0^1 f(x)dx &\text{by making a substitution $x=-x$}\\ & = \int_{0}^1 f(-x)dx + \int_0^1 f(x)dx &\text{switching the limits of integration}\\ & = \int_{0}^1 f(x)dx + \int_0^1 f(x)dx &\text{since $f(x) = f(-x)$}\\ & = 2\int_{0}^1 f(x)dx \end{align}
As to why you should do this rather than just try to integrate from $-1$ to $1$: It is kind of hard to integrate a function with an absolute value inside. However, from $0$ to $1$, the absolute value is nonnegative.