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

Q2 Find the area bounded by curves  \small (x-1)^2+y^2=1  and   \small x^2+y^2=1. 

Answers (1)

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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}

Posted by

Gautam harsolia

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