Mathematics - Area Under Curves Question with Solution | TestHub

Required area

$=\pi \times {\left(6\right)}^2-2\left({\int }_0^3\sqrt{9x}dx+{\int }_3^6\sqrt{36-x^2}dx\right)$

$=36\pi -6{\left[\frac{x^{3/2}}{3/2}\right]}_0^3-2{\left[\frac{x}{2}\sqrt{36-x^2}+18{\sin }^{-1}\frac{x}{6}\right]}_3^6$

$=36\pi -12\sqrt{3}-2\left(0-\frac{9\sqrt{3}}{2}+9\pi -3\pi \right)$

$=24\pi -3\sqrt{3} \mathrm{sq}.\mathrm{units}$