Find the volume formed by rotating the region bounded by $y = e^{-x} \sin x$, $x\ge 0$ about $y =0$.

Question

Find the volume formed by rotating the region bounded by $y = e^{-x} \sin x$, $x\ge 0$ about $y =0$.

I tried to graph this using Wolfram Alpha, but it didn't help. I don't know how to start or graph this.

Answer 1

Here is a graph from $0$ to $\pi$, the integral you already know from André ...

Answer 2

$$f(x) = e^{-x} \sin x$$ Your volume can be found using cylindrical coordinates :

\begin{align} \mathcal{V}\equiv\int\limits_0^\infty \int\limits_0^{2\pi}\int\limits_0^{f(x)} r dr d\theta dx &= 2\pi \int\limits_0^\infty\int\limits_0^{f(x)} r dr dx\\ &=\pi \int\limits_0^\infty f^2(x) dx =\pi \int\limits_0^\infty e^{-2x}\sin^2x dx = \pi \int\limits_0^\infty e^{-2x}\dfrac{1-\cos 2x}{2}dx\\ &= \dfrac{\pi}{2} \left\lbrace \int\limits_0^\infty e^{-2x}dx -\dfrac{1}{2} \int\limits_0^\infty e^{-2x}\left[e^{2ix} + e^{-2ix}\right]dx\right\rbrace\\ &= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} -\dfrac{1}{2} \left[\dfrac{e^{-2(1-i)x}}{-2(1-i)} + \dfrac{e^{-2(1+i)x}}{-2(1+i)}\right]_0^\infty\right\rbrace\\ &= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{4} \left(\dfrac{e^{2ix}}{1-i} + \dfrac{e^{-2ix}}{1+i}\right)\right]_0^\infty\right\rbrace\\ &= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{4} \left(\dfrac{e^{2ix}}{1-i} + \dfrac{e^{-2ix}}{1+i}\right)\right]_0^\infty\right\rbrace\\ &=\dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{8} \left((1+i)e^{2ix} + (1-i)e^{-2ix}\right)\right]_0^\infty\right\rbrace\\ &=\dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} +\left[\dfrac{e^{-2x}}{4} \left(\cos 2x - \sin 2x\right)\right]_0^\infty\right\rbrace\\ &= \dfrac{\pi}{2} \left\lbrace \dfrac{1}{2} - \dfrac{1}{4}\right\rbrace\\ &= \dfrac{\pi}{8} \end{align}