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