ক্ষেত্রফল নির্ণয়

The area enclosed between the parabolas $y^{2} = 16x$ and $x^{2} = 16y$ is

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Both parabolas cut each other at $(0,0)$ and $(16,16)$
Area enclosed by these parabolas
$=\displaystyle\int_0^{16}4\sqrt x dx-\int_0^{16} \dfrac{x^2}{16}dx$

$=\Bigl[\dfrac{2\times 4\times x^{3/2}}{3}\Bigr]_0^{16}-\Big[\dfrac{x^3}{16\times 3}\Bigr]_0^{16}$

$=\dfrac{2\times 4^4}{3}-\dfrac{4^4}{3}=\dfrac{256}{3}$ sq. units

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