Question

The base of a solid is the region in the first quadrant bounded above by the line y = 5, below by y = sin−1(x), and to the right by the line x = 1. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid?

Answer 1

Each cross section is a square with side length equal to the vertical distance between the curves $y=5$ and $y=\sin^{-1}x$, with $0\le x\le1$, or $5-\sin^{-1}x$. Then each section has an area of $(5-\sin^{-1}x)^2$, so the volume of this solid is

$\displaystyle\int_0^1(5-\sin^{-1}x)^2\,\mathrm dx=\boxed{33-5\pi+\frac{\pi^2}4}$

For computing the integral, consider the substitution

$u=5-\sin^{-1}x\implies x=\sin(5-u)\implies\mathrm dx=-\cos(5-u)\,\mathrm du$

Then the integral becomes

$\displaystyle\int_5^{5-\pi/2}u^2(-\cos(5-u))\,\mathrm du$

$=\displaystyle\int_{5-\pi/2}^5u^2\cos(u-5)\,\mathrm du$

Integrate by parts twice; for the first round, take

$f=u^2\implies\mathrm df=2u\,\mathrm du$

$\mathrm dg=\cos(u-5)\,\mathrm du\implies g=\sin(u-5)$

$\implies\displaystyle u^2\sin(u-5)\bigg|_{5-\pi/2}^5-2\int_{5-\pi/2}^5u\sin(u-5)\,\mathrm du$

For the second round, take

$f=u\implies\mathrm df=\mathrm du$

$\mathrm dg=-\sin(u-5)\,\mathrm du\implies g=\cos(u-5)$

$\implies\displaystyle\bigg(u^2\sin(u-5)+u\cos(u-5)\bigg)\bigg|_{5-\pi/2}^5-\int_{5-\pi/2}^5\cos(u-5)\,\mathrm du$

and the rest is trivial.