Question

The region in the first quadrant bounded by the x-axis, the line x = ln(π), and the curve y = sin(e^x) is rotated about the x-axis. What is the volume of the generated solid?

Answer 1

First, it would be good to know that the area bounded by the curve and the x-axis is convergent to begin with.

$\displaystyle\int_{-\infty}^{\ln\pi}\sin(e^x)\,\mathrm dx$

Let $u=e^x$, so that $\mathrm dx=\dfrac{\mathrm du}u$, and the integral is equivalent to

$\displaystyle\int_{u=0}^{u=\pi}\frac{\sin u}u\,\mathrm du$

The integrand is continuous everywhere except $u=0$, but that's okay because we have $\lim\limits_{u\to0^+}\frac{\sin u}u=1$. This means the integral is convergent - great! (Moreover, there's a special function designed to handle this sort of integral, aptly named the "sine integral function".)

Now, to compute the volume. Via the disk method, we have a volume given by the integral

$\displaystyle\pi\int_{-\infty}^{\ln\pi}\sin^2(e^x)\,\mathrm dx$

By the same substitution as before, we can write this as

$\displaystyle\pi\int_0^\pi\frac{\sin^2u}u\,\mathrm du$

The half-angle identity for sine allows us to rewrite as

$\displaystyle\pi\int_0^\pi\frac{1-\cos2u}{2u}\,\mathrm du$

and replacing $v=2u$, $\dfrac{\mathrm dv}2=\mathrm du$, we have

$\displaystyle\frac\pi2\int_0^{2\pi}\frac{1-\cos v}v\,\mathrm dv$

Like the previous, this require a special function in order to express it in a closed form. You would find that its value is

$\dfrac\pi2(\gamma-\mbox{Ci}(2\pi)+\ln(2\pi))$

where $\gamma$ is the Euler-Mascheroni constant and $\mbox{Ci}$ denotes the cosine integral function.