The region $R$ is bounded by the $x$-axis, the vertical lines $x = \frac 12$ and $x = a$ for some $a > 1$; and the graph of $y = \frac 1\pi(e^{x^2-x})$. Find, in terms of $a$, the volume of the solid obtained by rotating $R$ about the line $x = \frac 12$ by $2\pi$ radian.
Hi everyone, how do i solve this question using cylindrical shell method? May I kindly get the workings ? Thanks in advance!
Define $\tau = 2\pi$ and $e(x) = e^x$. The radius from the axis of rotation ($x = 1/2$) is $r$ and the height of the cylinder is $y$. The volume of the solid is given by
\begin{align*} \int_0^{a - 1/2} \tau r y \, dr &= \int_{1/2}^{a} \tau(x - 1/2) \frac{e(x^2 - x)}{\tau/2} \, dx \\ &= \int_{1/2}^a (2x - 1) e(x^2 - x) \, dx. \end{align*} Consider $u = x^2 - x$. Then, $du = (2x - 1) dx$. The endpoints $x = 1/2$ and $x = a$ are substituted by $u = -1/4$ and $u = a^2 - a$. Hence, \begin{align*} \int_{1/2}^a (2x - 1) e(x^2 - x) \, dx &= \int_{-1/4}^{a^2 - a} e(u) \, du \\ &= e(u) \bigg\vert_{u = -1/4}^{u = a^2 - a} \\ &= e(a^2 - a) - e(-1/4) \\ &:= e^{a^2 - a} - e^{-1/4} \\ &= e^{a(a - 1)} - \frac{1}{\sqrt[4]{e}}. \end{align*} Let me know if this was helpful.