# What is the volume of the solid produced by revolving f(x)=sinx, x in [0,pi] around the x-axis?

Apr 20, 2016

$V = {\pi}^{2} / 2$

#### Explanation: We have drawn the given expression $f \left(x\right) = \sin x$, for $x = 0$ to $x = \pi$. When this expression is revolved around $x$ axis through ${360}^{\circ}$ we have solids of revolution. At each point located on the graph, the cross-section of the solid, parallel to the $y$-axis, will be a circle of radius $y$.

Let us consider a thin disc of thickness $\delta x$, at a distance $x$ from the origin, with its face nearest to the $y$-axis. The radius of the other circular face of the disc will be y+δy.
The disc is not a perfect cylinder. It will become one when δx, and hence δy->0.

Thus we approximate the disc with a cylinder of thickness, δx and radius $y$.
The volume δV of the disc is then given by the volume of a cylinder, πr^2h, so that

δV=πy^2δx.
We have ignored the volume of part of the disc which is $\propto \left(\delta x \times \delta y\right)$, both being infinitesimally small.

So the volume $V$ of the solid of revolution is given by the integral of volume expression over the limits of interest as both δx->0 and δy->0.
V=int_a^b πy^2dx
Inserting values given in the problem we obtain
V=int_0^pi π(sinx)^2.dx
or V=int_0^pi πsin^2x.dx .....(1)
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To find the $\int {\sin}^{2} x . \mathrm{dx}$
From trigonometric identity ${\sin}^{2} x = \frac{1}{2} \left(1 - \cos 2 x\right)$
$\therefore$ we need to find $\int \frac{1}{2} \left(1 - \cos 2 x\right) . \mathrm{dx}$
$\implies \int \frac{1}{2} \mathrm{dx} - \frac{1}{2} \int \cos 2 x . \mathrm{dx}$
Integrating first term and for the second term
let $u = 2 x$. Differentiating both sides and inserting in the equation
$\mathrm{du} = 2 \mathrm{dx}$

$\implies \frac{x}{2} - \frac{1}{4} \int \cos u . \mathrm{du}$
$\implies \frac{x}{2} - \frac{1}{4} \int \cos u . \mathrm{du}$
$\implies \frac{x}{2} - \frac{1}{4} \sin u$, substituting back to $x$
$\implies \frac{x}{2} - \frac{1}{4} \sin 2 x$
We have ignored the constant of integration as we are eventually dealing with definite integral.
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Substituting the value of integral in (1)
$V = \pi {\left[\frac{x}{2} - \frac{1}{4} \sin 2 x\right]}_{0}^{\pi}$
$V = \pi \left[\left(\frac{\pi}{2} - \frac{1}{4} \sin 2 \pi\right) - \left(\frac{0}{2} - \frac{1}{4} \sin \left(2 \times 0\right)\right)\right]$, inserting the value of $\sin 2 \pi \mathmr{and} \sin 0 , \text{both} = 0$
$V = {\pi}^{2} / 2$