Question

The area of the region bounded by the curve y $=\frac{16-x^2}{4}$ and $y=se{c}^{-1}[-si{n}^{2}x],$ where [.] stands for the greatest integer function is:

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

Answer 1

Answer: (B)

$\frac{8}{3}(4-\pi {)}^{3/2}$

$[-si{n}^{2}x]=0or-1.Butse{c}^{-1}0$ is not defined .
$\therefore [-si{n}^{2}x]=-1.$
Hence the required area $=$ area between the parabola $x^2=-4(y-4)$ and the straight line
$y=se{c}^{-1}=\pi .$
Hence the required area$=$ 2area MAB
$2{\int }_{\pi }^{4}xdy=2{\int }_{\pi }^4\sqrt{16-4y}dy=4{\int }_{\pi }^4\sqrt{4-y}dy$
$=-4{\left\{\frac{(4-y{)}^{3/2}}{\frac{3}{2}}\right\}}_{\pi }^4=\frac{8}{3}(4-\pi {)}^{3/2}$