Chapter 6 - Applications of the Integral - 6.4 The Method of Cylindrical Shells - Exercises - Page 313: 62

$2\pi (1-c)$

Let us consider that $a= y=f(x)$. Now, we have: $\dfrac{dy}{dx}=-xy$ and $da=-x f(x) \ dx $ or, $-da=x f(x) dx$ $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(x) \ dx$ Now, $V=2\pi \int_{0}^{a} x f(x) \ dx \\= 2\pi \int_{1}^{c} (-1) \ da \\= 2\pi \int_c^1 da \\= 2\pi [a]_1^c \\= 2\pi (1-c)$