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]
Answer:
[tex]V=\frac{\pi^2}{3}+\frac{\pi\sqrt3}{8}[/tex]
Step-by-step explanation:
We are given that a region R is bounded by y= sin x and x- axis on the interval [tex][0,\frac{2\pi}{3}][/tex]
We have to write definite integral for the volume V of solid formed by R is revolved about x- axis and we have to find V.
Volume of solid is given by
[tex]V=\pi \int_{0}^{\frac{2\pi}{3}} sin^2 xdx[/tex]
We know that [tex]sin^2x=\frac{1-cos2x}{2}[/tex]
[tex]V=\frac{\pi}{2} \int_{0}^{\frac{2\pi}{3}} (1-cos 2x) dx[/tex]
[tex]V=\frac{\pi}{2}(x-\frac{sin2x}{2})^{\frac{2\pi}{3}}_{0}[/tex]
[tex]V=\frac{\pi}{2}(\frac{2\pi}{3}-\frac{1}{2}sin\frac{4\pi}{3})[/tex]
[tex]V=\frac{\pi}{2}(\frac{2\pi}{3}-\frac{1}{2}sin(\pi+\frac{\pi}{3}))[/tex]
[tex]V=\frac{\pi}{2}(\frac{2\pi}{3}+\frac{1}{2}sin\frac{\pi}{3})[/tex]
[tex] sin(\pi+\frac{\pi}{3})=-sin\frac{\pi}{3}=-\frac{\sqrt3}{2}[/tex]
[tex]V=\frac{\pi}{2}(\frac{2\pi}{3}+\frac{1}{2}\cdot\frac{\sqrt3}{2})[/tex]
[tex]V=\frac{\pi^2}{3}+\frac{\pi\sqrt3}{8}[/tex]
Hence, the volume of solid when R is revolved around the x- axis=[tex]V=\frac{\pi^2}{3}+\frac{\pi\sqrt3}{8}[/tex]
