If the area bounded by the curve 2y2 = 3x , lines x + y = 3, y = 0 and outside the circle (x - 3)2 + y2 = 2 is A , then 4(π + 4A) =

A=∫032 32x+△ABC− Area of sec tor =32x3232032+1232×32−12×2π4 =218-π4Then 4((π + 4A) = 42

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If the area bounded by the curve 2y² = 3x , lines x + y = 3, y = 0 and outside the circle (x - 3)² + y² = 2 is A , then 4( $π$ + 4A) =

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answer is 42.

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Detailed Solution

$A = \int_{0}^{\frac{3}{2}} \sqrt{\frac{3}{2}} \sqrt{x} + △ ABC - Area of \sec tor = {\sqrt{\frac{3}{2}} \frac{x^{\frac{3}{2}}}{\frac{3}{2}}|}_{0}^{\frac{3}{2}} + \frac{1}{2} (\frac{3}{2} \times \frac{3}{2}) - \frac{1}{2} \times 2 \frac{π}{4} = \frac{21}{8} - \frac{π}{4}$
Then 4(( $π$ + 4A) = 42

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