Question

Find the area of the region {$(x,y):y^2\le 6ax$ and $x^2+y^2\le 16a^2$} using method of integration.

Answer 1

We have $(x,y):y^2\le 6ax$ and $x^2+y^2\le 16a^2$

Consider $y^2=6ax$---- (1) and $x^2+y^2=16a^2$--- (2)

Curve (2) represents a circle centered at (0, 0) having radius of 4a.

Solving (1) and (2), $x^2+6ax-16a^2=0\Rightarrow (x+8a)(x-2a)=0$

Therefore, $x=2a$ and $x=-8a$

So, $x=2a,y=\pm 2\sqrt{3}a$ and point of intersection are $(2a,\pm 2\sqrt{3}a)$.

Now required area = $2\times ar\left(OAPMO\right)$

$=2[{\int }_0^{2a}{y}_{1}dx+{\int }_{2a}^{4a}{y}_{2}dx]$

$=2[\sqrt{6a}{\int }_0^{2a}\sqrt{x}dx+{\int }_{2a}^{4a}\sqrt{(4a{)}^2-x^2}dx]$

$=2\left(\frac{2}{3}\sqrt{6a}\right[x^{\frac{3}{2}}{]}_0^{2a}+{[\frac{x}{2}\sqrt{(4a{)}^2-x^2}+\frac{16a^2}{2}si{n}^{-1}\frac{x}{4a}]}_{2a}^{4a})$

$=\frac{4}{3}\sqrt{6a}\left[\right(2a{)}^{\frac{3}{2}}-0^{\frac{3}{2}}]+{[x\sqrt{(4a{)}^2-x^2}+16a^{2}si{n}^{-1}\frac{x}{4a}]}_{2a}^{4a}$

$=\frac{4}{3}\sqrt{6a}\sqrt{2a}\left(2a\right)+\left[\right(4a\times 0+16a^{2}si{n}^{-1}\left(1\right))-(2a\times 2\sqrt{3a}+16a^{2}si{n}^{-1}\frac{1}{2}\left)\right]$

$=\frac{4}{3}\sqrt{6a}\sqrt{2a}\left(2a\right)+\left[16a^2\right(\frac{\pi }{2}-(4\sqrt{3}{a}^2+16a^2(\frac{\pi }{6}\left)\right)]$

$=\frac{16\sqrt{3}}{3}{a}^2+[8\pi a^2-4\sqrt{3}-\frac{8\pi }{3}{a}^2]$

$=\frac{4\sqrt{3}}{3}{a}^2+\frac{16\pi }{3}{a}^2=\frac{4a^2}{3}(4\pi +\sqrt{3})$ Sq.units.