Question

Area of the region bounded by the curves $y=\frac{16-x^2}{4}$ and $y={\sec }^{-1}[-{\sin }^{2}x],$
$($where $[\cdot ]$ denotes greatest integer function $)$ is $($in sq. units$)$

• A. $\frac{8}{3}(4-\pi {)}^{\frac{3}{2}}$
• B. $2(4-\pi {)}^{\frac{3}{2}}$
• C. $3(4-\pi {)}^{\frac{3}{2}}$
• D. $\frac{7}{3}(4-\pi {)}^{\frac{3}{2}}$

Answer 1

The correct option is A $\frac{8}{3}(4-\pi {)}^{\frac{3}{2}}$

$-1\le -{\sin }^{2}x\le 0$
$\Rightarrow [-{\sin }^{2}x]=-1,0$
$y={\sec }^{-1}[-{\sin }^{2}x]$
$y={\sec }^{-1}0$ is not defined,
$y={\sec }^{-1}(-1)=\pi$

$\Rightarrow$ Point of intersection of $y=\frac{16-x^2}{4}$ & $y=\pi$ is
$x=\pm 2\sqrt{4-\pi }$
$\therefore$ Required Area
$=\underset{-2\sqrt{4-\pi }}{\overset{2\sqrt{4-\pi }}{\int }}(\frac{16-x^2}{4}-\pi )dx$
$=\frac{8}{3}(4-\pi {)}^{\frac{3}{2}}$