Question

Find the volume of the solid generated by rotating the region of the xy-plane between the line y=2, the curve y=sin(x)+1, for −π2≤x≤π2 about the line y=2.

Answer 1

This is best done with the disk method.

$\sin x=1$ for $x=\pm\frac\pi2$, so $\sin x+1=2$ at these points, and $|\sin x|\le1$ for all $x$ means $0\le\sin x+1\le2$.

Each disk has a radius of $2-(\sin x+1)=1-\sin x$, and contributes a volume of $\pi(1-\sin x)^2\,\mathrm\Delta x$ where $\Delta x$ is the height of each disk.

Then the volume of the solid is

$\displaystyle\pi\int_{-\pi/2}^{\pi/2}(1-\sin x)^2\,\mathrm dx$

$=\displaystyle\pi\int_{-\pi/2}^{\pi/2}(1-2\sin x+\sin^2x)\,\mathrm dx$

$=\pi\left(x+2\cos x+\dfrac{x-\sin x\cos x}2\right)\bigg|_{-\pi/2}^{\pi/2}=\boxed{\frac{3\pi^2}2}$