# Find the area bounded by curves

Question:

Find the area bounded by curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$

Solution:

The area bounded by the curves, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, is represented by the shaded area as

On solving the equations, $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$, we obtain the point of intersection as $\mathrm{A}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $\mathrm{B}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$.

It can be observed that the required area is symmetrical about $x$-axis.

$\therefore$ Area $\mathrm{OBCAO}=2 \times$ Area $\mathrm{OCAO}$

We join $\mathrm{AB}$, which intersects $\mathrm{OC}$ at $\mathrm{M}$, such that $\mathrm{AM}$ is perpendicular to $\mathrm{OC}$.

The coordinates of $M$ are $\left(\frac{1}{2}, 0\right)$.

$\Rightarrow$ Area $O C A O=$ Area $O M A O+$ Area $M C A M$

$=\left[\int_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} d x+\int_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} d x\right]$

$=\left[\frac{x-1}{2} \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}(x-1)\right]_{0}^{\frac{1}{2}}+\left[\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x\right]_{\frac{1}{2}}^{1}$

$=\left[-\frac{1}{4} \sqrt{1-\left(-\frac{1}{2}\right)^{2}}+\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}-1\right)-\frac{1}{2} \sin ^{-1}(-1)\right]+$

$\left[\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \sqrt{1-\left(\frac{1}{2}\right)^{2}}-\frac{1}{2} \sin ^{-1}\left(\frac{1}{2}\right)\right]$

$=\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi}{6}\right)-\frac{1}{2}\left(-\frac{\pi}{2}\right)\right]+\left[\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi}{6}\right)\right]$

$=\left[-\frac{\sqrt{3}}{4}-\frac{\pi}{12}+\frac{\pi}{4}+\frac{\pi}{4}-\frac{\pi}{12}\right]$

$=\left[-\frac{\sqrt{3}}{4}-\frac{\pi}{6}+\frac{\pi}{2}\right]$

$=\left[\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}\right]$

Therefore, required area $\mathrm{OBCAO}=2 \times\left(\frac{2 \pi}{6}-\frac{\sqrt{3}}{4}\right)=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right)$ units