Let Delta be the area of the region left{( x | vedclass

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(D) The region is bounded by the circle $x^2 + y^2 = 21$ and the parabola $y^2 = 4x$ for $x \geq 1$.First,find the intersection points: $x^2 + 4x - 21

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

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(D) The region is bounded by the circle $x^2 + y^2 = 21$ and the parabola $y^2 = 4x$ for $x \geq 1$.First,find the intersection points: $x^2 + 4x - 21 = 0 \implies (x+7)(x-3) = 0$. Since $x \geq 1$,we have $x = 3$.The area $\Delta$ is given by the sum of two integrals:$\Delta = 2 \int_1^3 2\sqrt{x} \, dx + 2 \int_3^{\sqrt{21}} \sqrt{21-x^2} \, dx$Evaluating the first part: $4 \left[ \frac{2}{3} x^{3/2} \right]_1^3 = \frac{8}{3} (3\sqrt{3} - 1) = 8\sqrt{3} - \frac{8}{3}$.Evaluating the second part: $2 \left[ \frac{x}{2} \sqrt{21-x^2} + \frac{21}{2} \sin^{-1} \left( \frac{x}{\sqrt{21}} \right) \right]_3^{\sqrt{21}}$$= 2 \left[ (0 + \frac{21}{2} \sin^{-1}(1)) - (\frac{3}{2} \sqrt{12} + \frac{21}{2} \sin^{-1} \left( \frac{3}{\sqrt{21}} \right)) \right]$$= 21 \left( \frac{\pi}{2} \right) - 6\sqrt{3} - 21 \sin^{-1} \left( \frac{3}{\sqrt{21}} \right) = \frac{21\pi}{2} - 6\sqrt{3} - 21 \sin^{-1} \left( \frac{\sqrt{3}}{\sqrt{7}} \right)$.Using $\sin^{-1} \left( \frac{\sqrt{3}}{\sqrt{7}} \right) = \cos^{-1} \left( \frac{2}{\sqrt{7}} \right) = \frac{\pi}{2} - \sin^{-1} \left( \frac{2}{\sqrt{7}} \right)$,we get:$\Delta = 8\sqrt{3} - \frac{8}{3} + 21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right) - 6\sqrt{3} = 2\sqrt{3} - \frac{8}{3} + 21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right)$.Thus,$\frac{1}{2} \left( \Delta - 21 \sin^{-1} \frac{2}{\sqrt{7}} \right) = \frac{1}{2} \left( 2\sqrt{3} - \frac{8}{3} \right) = \sqrt{3} - \frac{4}{3}$.

Let $\Delta$ be the area of the region $\left\{( x , y ) \in \mathbb{R} ^2: x ^2+ y ^2 \leq 21, y ^2 \leq 4 x , x \geq 1\right\}$. Then $\frac{1}{2}\left(\Delta-21 \sin ^{-1} \frac{2}{\sqrt{7}}\right)$ is equal to

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