Question

Using integration find the area of the region $\left\{\right(x,y):y^2\le 6ax\text{and}{x}^2+y^2\le 16a^2\}$

Answer 1

$\begin{array}{c}Area={\int }_0^{2a}\sqrt{6ax}dx+{\int }_{2a}^{4a}\sqrt{16a^2-x^2}dx\\ =\sqrt{6a}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{2a}+{\left[\frac{x\sqrt{16a^2-x^2}}{2}+\frac{16a^2}{2}{\sin }^{-1}\frac{x}{4a}\right]}_{2a}^{4a}\\ \sqrt{6a}\cdot \frac{2}{3}{\left(2a\right)}^{\frac{3}{2}}+\left[8a^2\frac{\pi }{2}-\left(a\sqrt{12a}+8a^2\times \frac{\pi }{6}\right)\right]\\ A=\frac{8}{\sqrt{3}}{a}^2+\frac{8\pi a^2}{3}-2\sqrt{3}{a}^2\\ =a^2\left(\frac{2\sqrt{3}+8\pi }{3}\right)uni{t}^2\\ Hence,solved.\end{array}$