The area (in sq. units) of the smaller part o | vedclass

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(C) The equation of the circle is $x^2+y^2=a^2$. The line is $x=\frac{a}{\sqrt{2}}$.Substituting $x=\frac{a}{\sqrt{2}}$ in the circle equation,we get

Vedclass · Accepted Answer

$\frac{a^2}{2}\left|\frac{\pi}{2}-1\right|$

Vedclass · Answer

(C) The equation of the circle is $x^2+y^2=a^2$. The line is $x=\frac{a}{\sqrt{2}}$.Substituting $x=\frac{a}{\sqrt{2}}$ in the circle equation,we get $\frac{a^2}{2}+y^2=a^2$,which implies $y^2=\frac{a^2}{2}$,so $y=\pm\frac{a}{\sqrt{2}}$.The required area is the area of the region bounded by the circle and the line $x=\frac{a}{\sqrt{2}}$ on the right side.Area $= 2 \int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2} dx$Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:Area $= 2 \left[ \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) \right]_{\frac{a}{\sqrt{2}}}^a$$= 2 \left[ (0 + \frac{a^2}{2}\sin^{-1}(1)) - (\frac{a}{2\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}} + \frac{a^2}{2}\sin^{-1}(\frac{1}{\sqrt{2}})) \right]$$= 2 \left[ \frac{a^2}{2}(\frac{\pi}{2}) - (\frac{a}{2\sqrt{2}}\cdot\frac{a}{\sqrt{2}} + \frac{a^2}{2}(\frac{\pi}{4})) \right]$$= 2 \left[ \frac{a^2\pi}{4} - \frac{a^2}{4} - \frac{a^2\pi}{8} \right]$$= 2 \left[ \frac{a^2\pi}{8} - \frac{a^2}{4} \right] = \frac{a^2\pi}{4} - \frac{a^2}{2} = \frac{a^2}{2}(\frac{\pi}{2}-1)$.

The area (in sq. units) of the smaller part of the circle $x^2+y^2=a^2$ cut off by the line $x=\frac{a}{\sqrt{2}}$ is

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