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Question

(A) The circle is $x^2 + y^2 = 2^2$, so the radius $r = 2$. The line is $x = 1$.The area of the smaller region is given by $A = 2 \int_{1}^{2} \sqrt{4

Vedclass · Accepted Answer

$\frac{4\pi}{3} - \sqrt{3}$

Vedclass · Answer

(A) The circle is $x^2 + y^2 = 2^2$, so the radius $r = 2$. The line is $x = 1$.The area of the smaller region is given by $A = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx$.Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$, we get:$A = 2 [\frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2})]_{1}^{2}$.$A = 2 [(\frac{2}{2} \sqrt{4 - 4} + 2 \sin^{-1}(1)) - (\frac{1}{2} \sqrt{4 - 1} + 2 \sin^{-1}(\frac{1}{2}))]$.$A = 2 [(0 + 2 \cdot \frac{\pi}{2}) - (\frac{\sqrt{3}}{2} + 2 \cdot \frac{\pi}{6})]$.$A = 2 [\pi - \frac{\sqrt{3}}{2} - \frac{\pi}{3}] = 2 [\frac{2\pi}{3} - \frac{\sqrt{3}}{2}] = \frac{4\pi}{3} - \sqrt{3}$.

The area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$ is...

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