The area out of the region bounded by $y^{2} = 4ax\ and\ \ x^{2} = 4ay,a 0$ in square unit is

Question

The area out of the region bounded by $y^{2} = 4ax\ and\ \ x^{2} = 4ay,a > 0$ in square unit is

Dr. Abubacker · Accepted Answer

(a)
The point of intersection of given curves are (0,0) and $(4a,4a)$.
$Required\ area = \int_{0}^{4a}{\left( 2\sqrt{a}\sqrt{x} - \frac{x^{2}}{4a} \right)dx}$
$= \left\lbrack 2\sqrt{a} \bullet \frac{x^{3/2}}{3/2} - \frac{x^{3}}{12a} \right\rbrack_{0}^{4a}$
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$= \frac{32a^{2}}{3} - \frac{16a^{2}}{3}$
$= \frac{16a^{2}}{3}\ sq\ units$

Q. The area out of the region bounded by $y^{2} = 4ax\ and\ \ x^{2} = 4ay,a > 0$ in square unit is

Explanation

Reviewed by Dr. Abubacker

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