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- None of theseB- 16-3- -C- 8 -3-8 -3D- 4 -3-

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Critical Point

Area of the r...

Question

Area of the region in which point p(x, y), {x > 0} lies; such that $y \leq \sqrt{16 - x^{2}}$ and $∣ ∣ t a n^{- 1} (\frac{y}{x}) ∣ ∣ \leq \frac{π}{3}$ 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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None of these

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Solution

The correct option is B $(\frac{8 π}{3} + 8 \sqrt{3})$

Required area is the area of shaded region (APOQ)
= area of $Δ O A Q$ + area of sector (OAP)
$= \frac{1}{2} \times 4 \times 4 \sqrt{3} + \frac{π (4 \times 4)}{6}$
$= (\frac{8 π}{3} + 8 \sqrt{3})$

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Area of the region in which point p(x, y), {x > 0} lies; such that y≤√16−x2 and ∣∣tan−1(yx)∣∣≤π3 is

The correct option is B (8π3+8√3) Required area is the area of shaded region (APOQ) = area of ΔOAQ + area of sector (OAP) =12×4×4√3+π(4×4)6 =(8π3+8√3)

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

The correct option is B $(\frac{8 π}{3} + 8 \sqrt{3})$

Required area is the area of shaded region (APOQ)
= area of $Δ O A Q$ + area of sector (OAP)
$= \frac{1}{2} \times 4 \times 4 \sqrt{3} + \frac{π (4 \times 4)}{6}$
$= (\frac{8 π}{3} + 8 \sqrt{3})$