Question

Find the area of the region bounded by the curves
$x^2+y^2=36$ and $y^2=9x$

Answer 1

$x^2+y^2=36$

$y^2=9x$

$x^2+9x=36$

$x^2+9x-36=0$

$x^2+12x-3x-36=0$

$\therefore x=-12,x=3$

Shaded region

$={\int }_0^{3}ydx+{\int }_3^{6}y\cdot dx$

$={\int }_0^3\sqrt{9x}dx+{\int }_3^6\sqrt{36-x^2}dx$

$=3{\int }_0^3\sqrt{x}dx+{\int }_3^6\sqrt{6^2-x^2}dx$

$=3{\left[\frac{x^{3/2}}{\frac{3}{2}}\right]}_0^3+{[\frac{x}{2}\sqrt{6-x^2}+\frac{6^2}{2}{\sin }^{-1}\left(\frac{x}{6}\right)]}_3^6$

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

$=\frac{3\times 2}{3}[3^{3/2}-0]+[0+18{\sin }^{-1}\left(\frac{6}{6}\right)-(\frac{3}{2}\sqrt{6^2-3^2}+18{\sin }^{-1}\left(\frac{3}{6}\right)]]$

$=2\sqrt{27}+18\frac{x}{2}-\frac{3}{2}\sqrt{27}+(-18\frac{x}{6})$

$=\frac{\sqrt{27}}{2}+9x-3x$

$=\frac{3\sqrt{3}}{2}+6x$ sq. units.