Question

Area enclosed by $x^2=4y$ and $y=\frac{8}{x^2+4}$ is ?

• A. $\pi -\frac{4}{3}$
• B. $\pi -\frac{1}{3}$
• C. $2\pi -\frac{4}{3}$
• D. $2\pi -\frac{1}{3}$

Answer 1

The correct option is C $2\pi -\frac{4}{3}$

$x^2=4y,y=\frac{8}{x^2+4}$

$y=\frac{8}{x^2+4}$
$\therefore$ Area required =
$2\times {\int }_0^2(\frac{8}{x^2+4}-\frac{x^2}{4})dx=2\pi -\frac{4}{3}$