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(A) The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the region bounded by the

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the region bounded by the circle and the line in the first and fourth quadrants.Since the circle is symmetrical about the $x$-axis,the total area is twice the area of the region in the first quadrant.Area $= 2 \int_{\frac{a}{\sqrt{2}}}^{a} y \, dx = 2 \int_{\frac{a}{\sqrt{2}}}^{a} \sqrt{a^{2}-x^{2}} \, dx$Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right)$:Area $= 2 \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}\right) \right]_{\frac{a}{\sqrt{2}}}^{a}$$= 2 \left[ \left( 0 + \frac{a^{2}}{2} \sin^{-1}(1) \right) - \left( \frac{a}{2\sqrt{2}} \sqrt{a^{2}-\frac{a^{2}}{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{1}{\sqrt{2}}\right) \right) \right]$$= 2 \left[ \frac{a^{2}}{2} \left(\frac{\pi}{2}\right) - \left( \frac{a}{2\sqrt{2}} \cdot \frac{a}{\sqrt{2}} + \frac{a^{2}}{2} \left(\frac{\pi}{4}\right) \right) \right]$$= 2 \left[ \frac{a^{2}\pi}{4} - \frac{a^{2}}{4} - \frac{a^{2}\pi}{8} \right] = 2 \left[ \frac{2a^{2}\pi - a^{2}\pi}{8} - \frac{a^{2}}{4} \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} \left( \frac{\pi}{2} - 1 \right)$.

Find the area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$.

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