Question

The region under the curve with equation y = 1/(2x-1) is rotated through four right angles about the x-axis to form a solid. Find the volume of the solid between x=3 and x = 8.

Answer 1

Using the disk method, the volume of the solid is

$\displaystyle \pi \int_3^8 \left(\dfrac1{2x-1}\right)^2 \, dx = \pi \int_3^8 \frac{dx}{(2x-1)^2}$

Substitute $u = 2x-1$ and $du=2\,dx$.

$\displaystyle \pi \int_3^8 \frac{dx}{(2x-1)^2} = \frac\pi2 \int_5^{15} \frac{du}{u^2} \\\\ ~~~~~~~~ = -\frac\pi2 \frac1u \bigg|_5^{15} \\\\ ~~~~~~~~ = -\frac\pi2 \left(\frac1{15} - \frac15\right) = \boxed{\frac\pi{15}}$