If the area bounded by the curve 2 y 2 = 3 x , lines x + y = 3 , y = 0 and outside the circle ( x - 3 ) 2 + y 2 = 2 is A, then 4 ( + 4 A ) is equal to _____ . [2023]

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Solution

Q.
If the area bounded by the curve $2 y^{2} = 3 x,$ lines $x + y = 3, y = 0$ and outside the circle ${(x - 3)}^{2} + y^{2} = 2$ is A, then $4 (π + 4 A)$ is equal to _____ . [2023]

Ans.
(42)

$Given, y^{2} = \frac{3 x}{2}, x + y = 3, y = 0, and {(x - 3)}^{2} + y^{2} = 2$

$\Rightarrow 2 y^{2} = 3 (3 - y)$

$\Rightarrow 2 y^{2} + 3 y - 9 = 0 \Rightarrow 2 y^{2} - 3 y + 6 y - 9 = 0$

$\Rightarrow (2 y - 3) (y + 3) = 0 \Rightarrow y = \frac{3}{2}, - 2$

$Required area, A = \int_{0}^{\frac{3}{2}} ((3 - y) - \frac{2 y^{2}}{3}) d y - \frac{π}{8} (2)$

$= {(3 y - \frac{y^{2}}{2} - \frac{2 y^{3}}{9})}_{0}^{3 / 2} - \frac{π}{4}$ $= 3 \times \frac{3}{2} - \frac{9}{8} - \frac{2}{9} (\frac{27}{8}) - \frac{π}{4}$

$= \frac{9}{2} - \frac{9}{8} - \frac{3}{4} - \frac{π}{4}$

$∴$ $4 A + π = 4 [\frac{9}{2} - \frac{9}{8} - \frac{3}{4}] = \frac{21}{2} = 10.50$

$∴$ $4 (4 A + π) = 42$

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