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 xB- 16-3-8 -3- 4 -3-D- -3-

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Area of the r...

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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} x)

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(\frac{16}{3} + 8 \sqrt{3})

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(4 \sqrt{3} - π)

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(\sqrt{3} - π)

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Similar questions

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

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

p (x, y), {x > 0}

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 :

The area bounded by the curve $(y - {sin}^{- 1} x)^{2} = x - x^{2}$ is

(x, y)

x y \leq 8, 1 \leq y \leq x^{2}

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