Compute the area of the figure which lies in the first quadrant inside the curve $x^{2} + y^{2} = 3 a^{2}$ & is bounded by the parabola $x^{2} = 2 a y$ & $y^{2} = 2 a x (a > 0)$ .

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Solution

Intersection points
$x^{2} = 2 a y$
$y = 2 a x$
$\Rightarrow$ $x = \frac{y}{2 a}$
$\frac{y^{2}}{4 a^{2}} = 2 a y$
$y = 8 a^{3}$
$x = 4 a^{2}$
Hence points $(0, 0)$ and $(4 a^{2}, 8 a^{3})$
Area $= \int_{0}^{4 a^{2}} \sqrt{2 a} d x - \frac{1}{2 a} \int_{0}^{4 a^{2}} x^{2} d x$
$=$ $[\sqrt{2 a}] [\frac{x \frac{3}{2}}{\frac{3}{2}}] \int_{0}^{4 a^{2}} - \frac{1}{2 a} [\frac{x^{3}}{3}] \int_{0}^{4 a^{2}}$
$=$ $[\sqrt{2 a}] \frac{2}{3} 8 a^{3} - \frac{1}{2 a} \frac{64 a^{6}}{3}$
$=$ $\frac{8 (2)^{3 / 2} a^{7 / 2}}{3} - \frac{32 a^{5}}{3}$
$\Rightarrow$ $\frac{16}{3} [\sqrt{2} a \frac{7}{2} - 2 a^{5}]$

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