#### Draw a rough sketch of the region $(x, y) : y^2 \leq 6ax\: \: and \: \: x^2 + y^2 \leq 16a^2$. Also find the area of the region sketched using method of integration.

$\text { The region }\left\{(x, y): y^{2} \leq 6 a x \text { and } x^{2}+y^{2} \leq 16 a^{2}\right\}$

By solving the equations:

$y^{2} \leq 6 a x \: \: and\: \: x^{2}+y^{2} \leq 16 a^{2}$

Through substituting for $\mathrm{y}^{2}$
$\\\Rightarrow x^{2}+6 a x=16 a^{2}\\ \Rightarrow(x-2 a)(x+8 a)=0\\ \therefore x=2 a\\ [ as x=-8 a is not possible ]\\$

Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by $\mathrm{A}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ or $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx} .$

[By the symmetry of the image w.r.t x axis]

$\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right] \\$

$\text { Required area }=2\left[\int_{0}^{2 \mathrm{a}} \sqrt{6 \mathrm{ax}} \mathrm{d} \mathrm{x}+\int_{2 \mathrm{a}}^{4 \mathrm{a}} \sqrt{(4 \mathrm{a})^{2}-\mathrm{x}^{2}} \mathrm{dx}\right]$

$\\=2\left[\sqrt{6 a}\left[\frac{2 x^{\frac{3}{2}}}{3}\right]_{0}^{2 a}+\left[\frac{x \sqrt{(4 a)^{2}-x^{2}}}{2}+\frac{(4 a)^{2}}{2} \sin ^{-1}\left(\frac{x}{4 a}\right)\right]_{2 a}^{4 a}\right] \\ =2\left(\frac{8}{3} \sqrt{3} a^{2}+4 \pi a^{2}-2 \sqrt{3} a^{2}-\frac{4 a^{2} \pi}{3}\right) \\ =\frac{4}{3} a^{2}[\sqrt{3}+4 \pi] \text { sq.units } \\$