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Q.

The area bounded by the circle $x^{2} + y^{2} = 16,$ and the ordinates $x = 0, x = 2$ is

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a

$2 \sqrt{3} + \frac{4 π}{3}$

b

$4 \sqrt{3} + \frac{8 π}{3}$

c

$2 \sqrt{3} - \frac{4 π}{3}$

d

$4 \sqrt{3} - \frac{8 π}{3}$

answer is B.

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Detailed Solution

$x^{2} + y^{2} = 16$

$R . A = 2 \int_{0}^{2} \sqrt{16 - x^{2}} d x$

$= 2 [2 \sqrt{3} + \frac{4 π}{3}] = 4 \sqrt{3} + \frac{8 π}{3}$

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