Let ∆be the area of the region (x,y)∈ℝ2:x2+y2≤21,y2≤4x,x≥1. Then 12Δ−21sin−1⁡27 is equal to

x2+y2≤21,y2≤4x,x≥1x2+4x≤21x2+4x−21=0(x+7)(x−3)=0x=3,−7∆=2∫13 2xdx+2∫321 21−x2dx=4⋅23x3213+2x221−x2+212sin−1⁡x21321=83[33−1]+0+21π2−3⋅23+21sin−1⁡321=83−13+21π2−63−21sin−1⁡37=83−83+21π2−63−21sin−1⁡37=23−83+21π2−21sin−1⁡37=23−83+21sin−1⁡2712Δ−21sin−1⁡27=3−43

Let ∆be the area of the region (x,y)∈ℝ2:x2+y2≤21,y2≤4x,x≥1. Then 12Δ−21sin−1⁡27 is equal to

x2+y2≤21,y2≤4x,x≥1x2+4x≤21x2+4x−21=0(x+7)(x−3)=0x=3,−7∆=2∫13 2xdx+2∫321 21−x2dx=4⋅23x3213+2x221−x2+212sin−1⁡x21321=83[33−1]+0+21π2−3⋅23+21sin−1⁡321=83−13+21π2−63−21sin−1⁡37=83−83+21π2−63−21sin−1⁡37=23−83+21π2−21sin−1⁡37=23−83+21sin−1⁡2712Δ−21sin−1⁡27=3−43

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

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

$2 \sqrt{3} - \frac{1}{3}$

$\sqrt{3} - \frac{2}{3}$

$2 \sqrt{3} - \frac{2}{3}$

answer is C.

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$\begin{array}{l} x^{2} + y^{2} \leq 21, y^{2} \leq 4 x, x \geq 1 \\ x^{2} + 4 x \leq 21 \\ x^{2} + 4 x - 21 = 0 \\ (x + 7) (x - 3) = 0 \\ x = 3, - 7 \\ ∆ = 2 \int_{1}^{3} 2 \sqrt{x} dx + 2 \int_{3}^{\sqrt{21}} \sqrt{21 - x^{2}} dx \\ = 4 \cdot \frac{2}{3} {(x^{\frac{3}{2}})}_{1}^{3} + 2 {[\frac{x}{2} \sqrt{21 - x^{2}} + \frac{21}{2} \sin^{- 1} \frac{x}{\sqrt{21}}]}_{3}^{\sqrt{21}} \\ = \frac{8}{3} [3 \sqrt{3} - 1] + [(0 + 21 \frac{π}{2}) - (3 \cdot 2 \sqrt{3} + 21 \sin^{- 1} \frac{3}{\sqrt{21}})] \\ = 8 (\sqrt{3} - \frac{1}{3}) + 21 \frac{π}{2} - 6 \sqrt{3} - 21 \sin^{- 1} (\frac{\sqrt{3}}{\sqrt{7}}) \\ = 8 \sqrt{3} - \frac{8}{3} + 21 \frac{π}{2} - 6 \sqrt{3} - 21 \sin^{- 1} (\frac{\sqrt{3}}{\sqrt{7}}) \\ = 2 \sqrt{3} - \frac{8}{3} + 21 \frac{π}{2} - 21 \sin^{- 1} \frac{\sqrt{3}}{\sqrt{7}} \\ = 2 \sqrt{3} - \frac{8}{3} + 21 \sin^{- 1} \frac{2}{\sqrt{7}} \\ \frac{1}{2} (Δ - 21 \sin^{- 1} \frac{2}{\sqrt{7}}) = \sqrt{3} - \frac{4}{3} \end{array}$