Question

Use a double integral to prove that the area of the circle with radius $R$and equation $r=2R\sin \theta$is$\pi R^2$.

Short answer

The area of the circle is $A=\pi R^2$

Step-by-step solution

Given information

The objective of this problem is to use double integral to prove that the area of the circle with radius $R$and equation $r=2R\sin \theta$is$\pi R^2$.

Calculation

Draw the circle

Plot of $r=2R\sin \theta$

Given circle is symmetrical about the horizontal axis. Therefore area of circle in polar form can be expressed as the twice of area of upper half circle.

$A=2{\int }_a^{a_j}{\int }_n^{n_2}rdrd\theta$

Here, ${\theta }_1=0,{\theta }_2=\frac{\pi }{2}$and $r_1=0,r_2=r$$A=2{\int }_0^{\pi /2}{\int }_0^{r-2\pi \sin \theta }rdrd\theta$

Integrate with respect to $r$first

$A=2{\int }_0^{\pi /2}{\left[\frac{r^2}{2}\right]}_0^{2R\sin \theta }d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

$A=2{\int }_0^{x/2}\left[\frac{(2R\sin \theta {)}^2-0}{2}\right]$

$$\begin{gathered} A=2R^2{\int }_0^{\pi /2}\left[2{\sin }^2\theta \right]d\theta \\ A=2R^2{\int }_0^{\pi /2}[1-\cos 2\theta ]d\theta \end{gathered}$$

Integrate with respect to $\theta$

$$\begin{gathered} A=2R^2{\left[\theta -\frac{1}{2}\sin 2\theta \right]}_0^{x/2}\left[\int \cos xdx=\sin x+C\right] \\ A=2R^2\left[\frac{\pi }{2}-\frac{1}{2}\sin \pi -0\right] \\ A=\pi R^2 \end{gathered}$$

Thus, the area of the circle is

$A=\pi R^2$