Answer to Question #348506 in Calculus for Migz

Question

A curve has an equation of y=sin x. If the area of the curve

y=sin x from x=0 to X=pi is revolved about the y-axis, what is the

volume generated?

Answer 1

$$V=\displaystyle\int_{0}^{\pi}2\pi x\sin xdx$$

$$\int x\sin xdx=-x\cos x+\int \cos x dx$$

$$=-x\cos x+\sin x +C$$

$$V=\displaystyle\int_{0}^{\pi}2\pi x\sin xdx$$

$$=[2\pi(-x\cos x+\sin x)]\begin{matrix} \pi\\ 0 \end{matrix}$$

$$=2\pi (\pi+0-0)=2\pi^2({units}^3)$$