The region R is bounded by y = sin x and the x-axis on the interval [0, 2Ο€/3]. Write a definite integral for the volume V of the solid formed when R is revolved around the x-axis and then find V. Type pi for Ο€ if needed in the integral limits.

Accepted Solution

The integral is[tex]\displaystyle\pi\int_0^{2\pi/3}\sin^2x\,\mathrm dx[/tex]Recall the double angle identity,[tex]\sin^2x=\dfrac{1-\cos(2x)}2[/tex]Then the volume is[tex]\displaystyle\frac\pi2\int_0^{2\pi/3}(1-\cos(2x))\,\mathrm dx=\frac\pi2\left(x-\frac12\sin(2x)\right)\bigg|_0^{2\pi/3}[/tex][tex]=\displaystyle\frac\pi2(\frac{2\pi}3-\frac12\sin\frac{4\pi}3\right)[/tex][tex]=\boxed{\dfrac{\pi^2}3+\dfrac{\pi\sqrt3}8}[/tex]