Problem Statement
The curve $C$ is defined by $y = \ln(2x - 1)$ for $x > 0.5$. The region $R$ is bounded by $C$, the lines $y = 0$, $y = 2$, and the $y$-axis.
Find the exact volume of the solid generated when $R$ is rotated through $2\pi$ radians about the $y$-axis.
Verified Solution & Marking Scheme
Invert to express x in terms of y
$$y = \ln(2x - 1) \implies e^y = 2x - 1 \implies x = \frac{e^y + 1}{2}$$
Set up volume of revolution integral
$$V = \pi \int_0^2 x^2 \, dy = \pi \int_0^2 \left( \frac{e^y + 1}{2} \right)^2 \, dy = \frac{\pi}{4} \int_0^2 (e^{2y} + 2e^y + 1) \, dy$$
Integrate term by term
$$\int_0^2 (e^{2y} + 2e^y + 1) \, dy = \left[ \frac{1}{2}e^{2y} + 2e^y + y \right]_0^2 = \left( \frac{1}{2}e^4 + 2e^2 + 2 \right) - \left( \frac{1}{2} + 2 + 0 \right) = \frac{1}{2}e^4 + 2e^2 - \frac{1}{2}$$
Calculate exact volume
$$V = \frac{\pi}{4} \left( \frac{e^4 + 4e^2 - 1}{2} \right) = \frac{\pi(e^4 + 4e^2 - 1)}{8}$$