Area of the region in which point p x - y - x -0 lies- such that y -16- x 2 and -tan -1 y - x - -3 isA- 16-3-B- 4 -3-C- 8 -3-8 -3D- -3-

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Area between x=g(y) and y Axis

Area of the r...

Question

Area of the region in which point $p (x, y), x > 0$ lies: such that $y \leq \sqrt{16 - x^{2}} a n d ∣ ∣ t a n^{- 1} (\frac{y}{x}) ∣ ∣ \leq \frac{π}{3}$ is

$(\frac{16}{3} π)$

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$(\frac{8 π}{3} + 8 \sqrt{3})$

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$(4 \sqrt{3} - π)$

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$(\sqrt{3} - π)$

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Solution

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y \leq \sqrt{16 - x^{2}}

∣ ∣ t a n^{- 1} (\frac{y}{x}) ∣ ∣ \leq \frac{π}{3}

The area (in sq. units) of the region ${(x, y) : y^{2} \geq 2 x a n d x^{2} + y^{2} \leq 4 x, x \geq 0, y \geq 0}$ is :

{(x, y) : \frac{2}{1 + x^{2}} \geq y \geq x^{2}}

y = x^{2}

y = \frac{2}{1 + x^{2}}

p (x)

\frac{p (x + 1)}{p (x)} = \frac{x^{2} + x + 1}{x^{2} - x + 1}

p (2) = 3

\int_{0}^{1} {tan}^{- 1} (p (x)) \cdot {tan}^{- 1} \sqrt{\frac{x}{1 - x}} d x

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