# 010. Shell Integration

#### Given $f(x) = \pi x^2 \sin \pi x^2$, calculate the volume $V$ obtained by rotating the area between $f(x) (0 \leq x \leq 1)$ and $x$-axis around $y$-axis.

After drawing $f(x)$, it is found that $f(x)$ can have two intersections with a line that is perpendicular to the rotation axis. So, it is hard to apply the basic volume calculation method. In this case, shell integration is much more useful. As shown below, the volume $\Delta V$ obtained by rotating the area between $f(x) (t \leq x \leq t + \Delta t)$ and $x$-axis around $y$-axis can be approximated as follows.

$\begin{aligned} \Delta V &\approx \pi \{ (t + \Delta t)^2 - t^2 \} f(t) = \pi \{ 2t\Delta t + (\Delta t)^2 \} f(t) \\\\ \implies \frac{\Delta V}{\Delta t} &= \pi (2t + \Delta t) f(t) \implies \lim_{\Delta t \to 0} \frac{\Delta V}{\Delta t} = \frac{dV}{dt} = 2\pi t f(t) \end{aligned}$

Note that $\Delta V$ can be also approximated as $\pi \{ (t + \Delta t)^2 - t^2 \} f(t + \Delta t)$. In this case, however, the same result is induced. Therefore, $V$ is $\begin{aligned} V &= \int_{0}^{1} 2\pi t f(t) dt = \int_{0}^{1} 2\pi x f(x) dx = \int_{0}^{1} 2\pi^2 x^3 \sin \pi x^2 dx \\\\ &= \int_{0}^{\pi} u \sin u du \quad (u = \pi x^2, \; du = 2\pi x dx) \\\\ &= \left[ -u \cos u \right]_{0}^{\pi} + \int_{0}^{\pi} \cos u du = \pi \end{aligned}$