The area of the part of the circle $x^{2} + y^{2} = 8 a^{2}$ and the parabola $y^{2} = 2 a x$ through which positive X-axis passes is

4 a^{2} (\frac{3 π + 2}{3})

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2 a^{2} (\frac{3 π - 2}{3})

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(\frac{a^{2} (3 π + 2)}{3})

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(\frac{2 a^{2} (3 π + 2)}{3})

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Solution

The correct option is D $(\frac{2 a^{2} (3 π + 2)}{3})$

$x^{2} + y^{2} = 8 a^{2}, y^{2} = 2 a x \Rightarrow x^{2} + 2 a x - 8 a^{2} = 0$
$\Rightarrow (x + 4 a) (x - 2 a) = 0 \Rightarrow x = 2 a, - 4 a$
Required area = $2 \int_{0}^{2 a} \sqrt{2} \sqrt{a} \sqrt{x} d x + 2 ∣ ∣ \int_{0}^{2 a} \sqrt{8 a^{2} - x^{2}} ∣ ∣ d x$
$= 2 \sqrt{2} \sqrt{a} x^{\frac{3}{2}} \times \frac{2}{3}]_{0}^{2 a} + 2 ∣ ∣ ∣ {[\frac{x}{2} \sqrt{8 a^{2} - x^{2}} + \frac{8 a^{2}}{2} s i n^{- 1} (\frac{x}{2 \sqrt{2 a}})]}_{2 a}^{- 1} ∣ ∣ ∣$
$= \frac{4 \sqrt{2}}{3} \sqrt{a} (2 a)^{\frac{3}{2}} + 2 (- \frac{29}{2} \sqrt{8 a^{2} - x^{2}} + 4 a^{2} (\frac{π}{2} - \frac{π}{4}))$
$= \frac{16}{3} a^{2} + 2 (- 2 a^{2} + π a^{2}) = \frac{4 a^{2} + + 6 π a^{2}}{3} = \frac{2 a^{2} (3 π + 2)}{3}$

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Similar questions

\frac{4}{3} (4 π - \sqrt{3})

\frac{4}{3} (4 π + \sqrt{3})

\frac{4}{3} (8 π - \sqrt{3})

\frac{4}{3} (8 π + \sqrt{3})

(1, \sqrt{3})

x^{2} + y^{2} = 4

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

y^{2} = 6 a x

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

y^{2} = 4 a (a - | x - a |); a > 0

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