Using integration, find the area of the region bounded by the curves $y^{2} = 4 a x$ and $x^{2} = 4 a y$ , where $a > 0$ .

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Solution

$y^{2} = 4 a x$
$(\frac{x^{2}}{4 x})^{2} = 4 a x$
$x^{4} = 64 a^{3} x$
$x (x^{2} - 64 a^{3}) = 0$
$∴ x = 0, x = 4 a$
$A x n = \int_{0}^{4 a} \sqrt{4 a x} d x - \int_{0}^{4 a} \frac{x^{2}}{4 a} d x$

$= \sqrt{4 a} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ - \frac{1}{49} {[\frac{x^{3}}{3}]}_{0}^{4 a}$

$= \sqrt{4 a} \times \frac{2}{3} \times (4 a)^{\frac{3}{2}} - \frac{1}{4 a} \times \frac{64 a^{3}}{3}$

$= 16 a^{2} \times \frac{2}{3} - \frac{16 a^{2}}{3} = \frac{16 a^{2}}{3}$

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