Question

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

Answer 1

Intersection points

$x^2=2ay$

$y=2ax$

$\Rightarrow$ $x=\frac{y}{2a}$

$\frac{y^2}{4a^2}=2ay$

$y=8a^3$

$x=4a^2$

Hence points $(0,0)$ and $(4a^2,8a^3)$

Area$={\int }_0^{4a^2}\sqrt{2a}dx-\frac{1}{2a}{\int }_0^{4a^2}{x}^{2}dx$

$=$ $\left[\sqrt{2a}\right]\left[\frac{x\frac{3}{2}}{\frac{3}{2}}\right]{\int }_0^{4a^2}-\frac{1}{2a}\left[\frac{x^3}{3}\right]{\int }_0^{4a^2}$

$=$$\left[\sqrt{2a}\right]\frac{2}{3}8a^3-\frac{1}{2a}\frac{64a^6}{3}$

$=$$\frac{8(2{)}^{3/2}{a}^{7/2}}{3}-\frac{32a^5}{3}$

$\Rightarrow$ $\frac{16}{3}[\sqrt{2}a\frac{7}{2}-2a^5]$