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Volume of a football

mark

Suppose we spin the region between the graph of y=\sin(x) and over the interval [0,\pi] about the x-axis We generate something that looks a bit like a football. Find the volume of that football.

mearing

Volume of a football

y=sin (x) [0,\pi]

\int_0^\pi\pi sin^2(x)dx

=\frac{\pi}{2}\int_0^\pi(1-cos(2x)dx

Let u=2x so that \frac{1}{2}du=dx

=\frac{\pi}{4}\int_0^{2\pi}(1-cos(u))du=\pi\frac{u}{4}-\pi\frac{sin(u)}{4}|_0^{2\pi}=\frac{\pi^2}{2}

knguyen3

V=\pi\int_0^\pi sin^2(x)dx=\pi\int_0^\pi \frac{1}{2}-\frac{cos(2x)}{2}dx=\pi(\frac{x}{2}-\frac{sin(2x)}{4})|_0^\pi=\frac{\pi^2}{2}

chowell1

The expression we are given is

y = sin(x)

on the interval [0,\pi]. We can approximate the area by using the formula for disks:

\int_0^{\pi}\pi sin^2(x)dx \ = \frac{\pi}{2}\int_0^{\pi}(1-cos(2x))dx \\

= \frac{\pi}{2}(\pi-sin(2\pi)-0-sin(0)) = \frac{\pi}{2}(\pi-0) = \frac{\pi^2}{2}