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

The correct option is A $\frac{a^2}{3}(6\pi -4)$



By Solving equation of the curves:
$\frac{x^2}{4a}=\frac{8a^3}{x^2+4a^2}$
$\Rightarrow x^4+4a^2{x}^2=32a^4$
$\Rightarrow (x^2-4a^2)(x^2+8a^2)=0$
$\Rightarrow x^2=4a^2\{\because x^2\ne -8\}$
$\Rightarrow x=\pm 2a$
Both are even function and points of intersection are symmetric along the $y$ axis.
$\{\because {\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx\}$
$\Rightarrow$ Area
$=2\underset{0}{\overset{2a}{\int }}(\frac{8a^3}{x^2+4a^2}-\frac{x^2}{4a})dx=2\times 8a^3\times \frac{1}{2a}{\left[{\tan }^{-1}\frac{x}{2a}\right]}_0^{2a}-\frac{2}{4a}\times {\left(\frac{x^3}{3}\right)}_0^{2a}$
$=8a^2{\tan }^{-1}\frac{2a}{2a}-\frac{1}{2a}\times \frac{8a^3}{3}$
$=8a^2\times \frac{\pi }{4}-\frac{4a^2}{3}$
$=\frac{a^2}{3}(6\pi -4)$