Question

The region R is bounded by y = sin x and the x-axis on the interval [0, 2π/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 π if needed in the integral limits.

Answer 1

The integral is

$\displaystyle\pi\int_0^{2\pi/3}\sin^2x\,\mathrm dx$

Recall the double angle identity,

$\sin^2x=\dfrac{1-\cos(2x)}2$

Then the volume is

$\displaystyle\frac\pi2\int_0^{2\pi/3}(1-\cos(2x))\,\mathrm dx=\frac\pi2\left(x-\frac12\sin(2x)\right)\bigg|_0^{2\pi/3}$

$=\displaystyle\frac\pi2(\frac{2\pi}3-\frac12\sin\frac{4\pi}3\right)$

$=\boxed{\dfrac{\pi^2}3+\dfrac{\pi\sqrt3}8}$

Answer 2

Answer:

$V=\frac{\pi^2}{3}+\frac{\pi\sqrt3}{8}$

Step-by-step explanation:

We are given that a region R is bounded by y= sin x and x- axis on the interval $[0,\frac{2\pi}{3}]$

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

$V=\pi \int_{0}^{\frac{2\pi}{3}} sin^2 xdx$

We know that $sin^2x=\frac{1-cos2x}{2}$

$V=\frac{\pi}{2} \int_{0}^{\frac{2\pi}{3}} (1-cos 2x) dx$

$V=\frac{\pi}{2}(x-\frac{sin2x}{2})^{\frac{2\pi}{3}}_{0}$

$V=\frac{\pi}{2}(\frac{2\pi}{3}-\frac{1}{2}sin\frac{4\pi}{3})$

$V=\frac{\pi}{2}(\frac{2\pi}{3}-\frac{1}{2}sin(\pi+\frac{\pi}{3}))$

$V=\frac{\pi}{2}(\frac{2\pi}{3}+\frac{1}{2}sin\frac{\pi}{3})$

$sin(\pi+\frac{\pi}{3})=-sin\frac{\pi}{3}=-\frac{\sqrt3}{2}$

$V=\frac{\pi}{2}(\frac{2\pi}{3}+\frac{1}{2}\cdot\frac{\sqrt3}{2})$

$V=\frac{\pi^2}{3}+\frac{\pi\sqrt3}{8}$

Hence, the volume of solid when R is revolved around the x- axis=$V=\frac{\pi^2}{3}+\frac{\pi\sqrt3}{8}$