2 Find the area bounded by curves and .

Name: Find the area bounded by curves (x – 1)^2 + y^2 = 1 and x^2 + y^2 = 1.
Uploaded: Wed, 2019-06-26 13:13
Description: Find the area bounded by curves (x – 1)^2 + y^2 = 1 and x^2 + y^2 = 1.

Find the area bounded by curves (x – 1)^2 + y^2 = 1 and x^2 + y^2 = 1.

2 Find the area bounded by curves $\small (x-1)^2+y^2=1$ and $\small x^2+y^2=1$ .

Answers (1)

Given curves are    $\small (x-1)^2+y^2=1$ and $\small x^2+y^2=1$

Point of intersection of these two curves are

$A = \left ( \frac{1}{2},\frac{\sqrt3}{2} \right )$ and    $B = \left ( \frac{1}{2},-\frac{\sqrt3}{2} \right )$

We can clearly see that the required area is symmetrical about the x-axis

Therefore,

Area of OBCAO = 2 $\times$ Area of OCAO

Now, join AB such that it intersects the x-axis at M and AM is perpendicular to OC

Coordinates of M = $\left ( \frac{1}{2},0 \right )$

Now,

Area OCAO = Area OMAO + Area CMAC

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

$=\left [- \frac{1}{4}\sqrt{1-(-\frac{1}{2})^2}+\frac{1}{2}\sin^{-1}(\frac{1}{2}-1)-0-\frac{1}{2}\sin^{-1}(-1) \right ]+\left [ 0+\frac{1}{2}\sin^{-1}(1)- \frac{1}{4}\sqrt{1-\left ( \frac{1}{2} \right )^2}-\frac{1}{2}\sin^{-1}\left ( \frac{1}{2} \right )\right ]$
$=\left [ -\frac{\sqrt3}{8}+\frac{1}{2}\left ( -\frac{\pi}{6} \right )-\frac{1}{2}\left ( -\frac{\pi}{2} \right ) \right ]+\left [ \frac{1}{2}\left ( \frac{\pi}{2} \right ) -\frac{\sqrt3}{8}-\frac{1}{2}\left ( \frac{\pi}{6} \right )\right ]$
   $= \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]+\left [ \frac{\pi}{6}-\frac{\sqrt3}{8} \right ]$
   $=2 \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]$
Now,
Area of OBCAO = 2 $\times$ Area of OCAO

$=2\times 2 \left [ -\frac{\sqrt3}{8}+\frac{\pi}{6} \right ]$
   $=\frac{2\pi}{3}-\frac{\sqrt3}{2}$

Therefore, the answer is $\frac{2\pi}{3}-\frac{\sqrt3}{2}$