The area of the portion of the circle $x^{2} + y^{2} = 64$ which is exterior to the parabola $y^{2} = 12 x$ , is

(8 π - \sqrt{3})

sq units

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

sq units

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\frac{16}{3} (8 π - \sqrt{3})

sq units

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None of the above

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Solution

The correct option is C $\frac{16}{3} (8 π - \sqrt{3})$ sq units
Required shaded area $=$

$= A r e a o f c i r c l e - 2 [\int_{0}^{4} 2 \sqrt{3} \sqrt{x} d x + \int_{4}^{8} \sqrt{64 - x^{2}} d x]$

$= 64 π - 2 [{(2 \sqrt{3} x^{3 / 2} \times \frac{2}{3})}_{0}^{4} + {(\frac{x}{2} \sqrt{64 - x^{2}} + \frac{64}{2} {sin}^{- 1} \frac{x}{8})}_{4}^{8}]$

$= 64 π - \frac{64}{\sqrt{3}} - 32 π + 16 \sqrt{3} + \frac{32 π}{3}$

$= \frac{128 π}{3} - \frac{16 \sqrt{3}}{3}$

$= \frac{16}{3} (8 π - \sqrt{3})$ sq units.

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\frac{16}{3}

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Choose the correct answer
The area of the circle $x^{2} + y^{2} = 16$ exterior to the parabola $y^{2} = 6 x$ is
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Q. The area (in sq. units) of the part of the circle

x^{2} + y^{2} = 36

, which is outside the parabola

y^{2} = 9 x,

is :

The area of the circle x² + y² = 16 exterior to the parabola y² = 6x is

Axes are coordinate axes. A and B are the ends of major axis and minor axis respectively.

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The area of the portion of the circle x2+y2=64 which is exterior to the parabola y2=12x, is

The correct option is C 163(8π−√3) sq unitsRequired shaded area = =Areaofcircle−2[∫402√3√xdx+∫84√64−x2dx]=64π−2[(2√3x3/2×23)40+(x2√64−x2+642sin−1x8)84]=64π−64√3−32π+16√3+32π3=128π3−16√33=163(8π−√3) sq units.

The area of the portion of the circle $x^{2} + y^{2} = 64$ which is exterior to the parabola $y^{2} = 12 x$ , is