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  • Find the area of the smaller part of the circle x^2 + y^2 = a^2 cut off by the line x=a/sqrt2.

Q : 7          Find the area of the smaller part of the circle \small x^2+y^2=a^2 cut off by the line 
                  \small x=\frac{a}{\sqrt{2}}

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Exercise 8.1
we need to find the area of smaller part of the circle
Now,
Area of ABCD = 2 X Area of ABC 
Area of ABC = \int_{\frac{a}{\sqrt2}}^{a} ydx= \int_{\frac{a}{\sqrt2}}^{a} \sqrt{a^2-x^2}dx= \left [ \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right ]^{a}_\frac{a}{\sqrt2}\\ \\
                                                                                   =\left [ \frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}\sin^{-1}\frac{a}{a}- \frac{a}{2\sqrt2}\sqrt{a^2-(\frac{a}{\sqrt2})^2}-\frac{a^2}{2}\sin^{-1}\frac{a}{a\sqrt2}\right ]
                                                                                   =\left [ \frac{a}{2}\sqrt{0}+\frac{a^2}{2}\sin^{-1}1- \frac{a}{2\sqrt2}\sqrt{\frac{a^2}{2}}-\frac{a^2}{2}\sin^{-1}\frac{1}{\sqrt2}\right ]
                                                                                    =\left [ 0+\frac{a^2}{2}\frac{\pi}{2}- \frac{a^2}{4}-\frac{a^2}{2}\frac{\pi}{4}\right ]
                                                                                    =\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )
Area of ABCD = 2 X Area of ABC
                       =2\times\frac{a^2}{4}\left ( \frac{\pi}{2}-1 \right )= \frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )
Therefore, the area of the smaller part of the circle is \frac{a^2}{2}\left ( \frac{\pi}{2}-1 \right )
                                                                                     
 

Posted by

Gautam harsolia

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