The area of the smaller segment cut off from | vedclass

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(B) The circle is $x^2 + y^2 = 3^2$,with radius $r = 3$. The line $x = 1$ cuts the circle. The area of the smaller segment is given by the integral of

Vedclass · Accepted Answer

$9\sec^{-1}(3) - \sqrt{8}$

Vedclass · Answer

(B) The circle is $x^2 + y^2 = 3^2$,with radius $r = 3$. The line $x = 1$ cuts the circle. The area of the smaller segment is given by the integral of $2y$ with respect to $x$ from $x = 1$ to $x = 3$.Area $= 2 \int_{1}^{3} \sqrt{9 - x^2} \, dx$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{9 - x^2} + \frac{9}{2} \sin^{-1}(\frac{x}{3}) ight]_{1}^{3}$$= \left[ x \sqrt{9 - x^2} + 9 \sin^{-1}(\frac{x}{3}) ight]_{1}^{3}$$= (3 \sqrt{9 - 9} + 9 \sin^{-1}(1)) - (1 \sqrt{9 - 1} + 9 \sin^{-1}(\frac{1}{3}))$$= (0 + 9 \cdot \frac{\pi}{2}) - (\sqrt{8} + 9 \sin^{-1}(\frac{1}{3}))$$= \frac{9\pi}{2} - \sqrt{8} - 9 \sin^{-1}(\frac{1}{3})$$= 9(\frac{\pi}{2} - \sin^{-1}(\frac{1}{3})) - \sqrt{8}$Since $\frac{\pi}{2} - \sin^{-1}(	heta) = \cos^{-1}(	heta)$ and $\cos^{-1}(\frac{1}{3}) = \sec^{-1}(3)$:Area $= 9 \sec^{-1}(3) - \sqrt{8}$.

The area of the smaller segment cut off from the circle $x^2 + y^2 = 9$ by the line $x = 1$ is

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