Find the area of the region bounded by the curves \left ( x-1 \right )^{2}+y^{2}= 1\: \: and\: \: x^{2}+y^{2}= 1, using integration.

 

 

 

 
 
 
 
 

Answers (1)

\left ( x-1 \right )^{2}+y^{2}= 1\: \left ( given \right )                    
\left ( x-1 \right )^{2}+\left ( y-0 \right )^{2}= 1                         
Thus,                                                                 
      center=(1,0)                                               
      radius=1   

x^{2}+y^{2}= 1\: \left ( given \right )
\left ( x-0 \right )^{2}+\left ( y-0 \right )^{2}= 1
Thus
 center=(0,0)
  radius=1
Area required = Area  OACB
First, we find Intersection
points A & B

x^{2}+y^{2}= 1\: ---(1)
\left ( x-1 \right )^{2}+y^{2}= 1\: ---(2)
From equation (1)
x^{2}+y^{2}= 1
y^{2}= 1-x^{2}
put y^{2}= 1-x^{2} in equation (2)
\left ( x-1 \right )^{2}+y^{2}= 1
\left ( x-1 \right )^{2}-x^{2}= 0
x^{2}-2x+1-x^{2}= 0
1= 2x
x= \frac{1}{2}
puting x= \frac{1}{2}  in equation (1)

x^{2}+y^{2}= 1
\left ( \frac{1}{2} \right )^{2}+y^{2}= 1
y^{2}= 1-\frac{1}{4}
y^{2}= \frac{3}{4}\: \: \: y= \frac{\pm \sqrt{3}}{2}
so y= \frac{\sqrt{3}}{2}\: \: ,\frac{-\sqrt{3}}{2}
so intersecting  points are
A= \left ( \frac{1}{2},\frac{\sqrt{3}}{2} \right )  &  B= \left ( \frac{1}{2},\frac{-\sqrt{3}}{2} \right )
Now finding area
Area= Area ACBD + Area OADB
Area ACBD
\because ACBD is symmetic about x-axis
so Area ACBD = 2\times Area ACD
            = 2\int_{\frac{1}{2}}^{1}ydx
\rightarrow equation of 1st circle
x^{2}+y^{2}= 1
y^{2}= 1-x^{2}
y= \pm \sqrt{1-x^{2}}
\therefore Area AOD is in 1st quadrant we to be positive value
so= y= \sqrt{1-\left ( x-1 \right )^{2}}
Hence
Area\: \: OADB=2\int_{0}^{\frac{1}{2}}ydx
= 2\int_{0}^{\frac{1}{2}}\sqrt{1-\left ( x-1 \right )^{2}dx}
putting t= x-1
Differentiating w.r.t x
\frac{dt}{dx}= 1                                    \frac{x}{t}\:\: \frac{0}{t= -1}\: \: \frac{\frac{-1}{2}}{t= \frac{-1}{2}}
dt= dx
so, = 2\int_{0}^{\frac{1}{2}}\sqrt{1-\left ( x-1 \right )^{2}dx}
      = 2\int_{0}^{\frac{1}{2}}\sqrt{1-t^{2}dt}
      = 2\left [ \frac{1}{2}\sqrt{1-t^{2}}+\frac{1^{2}}{2}\sin^{-1}\frac{t}{1} \right ]^{\frac{1}{2}}_{-1}

      = 2\left [ \frac{\left ( \frac{-1}{2} \right )}{2} \sqrt{1-\left ( \frac{-1}{2} \right )^{2}}+\frac{1^{2}}{2}\sin^{-1}\frac{\left ( \frac{-1}{2} \right )}{1}-\left [ \frac{\left ( -1 \right )}{2}\sqrt{1-\left ( -1 \right )^{2}} +\frac{1}{2}\sin^{-1}\left ( -1 \right )\right ]\right ]
= 2\left [ -\frac{1}{2}\sqrt{1-\frac{1}{4}+\frac{1}{2}}\times \frac{-\pi }{6}+\sqrt{0-\frac{1}{2}}\left ( \frac{-\pi }{2} \right )\right ]
= 2\left [ \frac{-1}{4}\sqrt{\frac{3}{4}}-\frac{\pi }{12}+\frac{\pi }{4} \right ]
\Rightarrow 2\left [ \frac{-\sqrt{3}}{4\times 2} +\frac{\pi }{4}-\frac{\pi }{12}\right ]
\Rightarrow 2\left [ \frac{-\sqrt{3}}{8}+\frac{3\pi -\pi }{12} \right ]\Rightarrow 2\left [ \frac{-\sqrt{3}}{8} +\frac{2\pi }{12}\right ]
\Rightarrow \frac{-\sqrt{3}}{4}+\frac{\pi }{3}
Area required = Area ACBD + Area OADB
= \frac{\pi }{3}-\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}+\frac{\pi }{3}
=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}



 

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