Question

Area bounded by $x^2-4ay$ and $y=\frac{8a^3}{x^2+4a^2}$ is

• A. $\frac{a^2}{3}(6\pi -4)$
• B. $\frac{\pi a^2}{3}$
• C. $\frac{a^2}{3}(6\pi +4)$
• D. 0

Answer 1

Answer: (D)

by equating $\frac{x^2}{4a}=\frac{8a^3}{x^2+4a^2}$
$\Rightarrow \pm 2a$
Area $=2{\int }_0^{2a}\left(\frac{8a^3}{x^2+4a^2}-\frac{x^2}{4a}\right)dx$