0
$\begingroup$

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.

  • 0
    You shouldn't get zero because your integrand is non-negative... which is expected because you don't want to get a zero surface area. In general, it is indeed useful to notice symmetries that might reduce calculations. In your case, the integrand is even so maybe you can exploit this. – 2017-02-23

1 Answers 1

2

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.