MCQ
The area in the first quadrant between ${x^2} + {y^2} = {\pi ^2}$ and $y = \sin x$ is
  • $\frac{{({\pi ^3} - 8)}}{4}$
  • B
    $\frac{{{\pi ^3}}}{4}$
  • C
    $\frac{{({\pi ^3} - 16)}}{4}$
  • D
    $\frac{{({\pi ^3} - 8)}}{2}$

Answer

Correct option: A.
$\frac{{({\pi ^3} - 8)}}{4}$
a
(a) Area of the circle in first quadrant is $\frac{{\pi ({\pi ^2})}}{4}$

$i.e.,$ $\frac{{{\pi ^3}}}{4}$.

Also area bounded by curve $y = \sin x$ and $x -$ axis is $2$ sq. unit. 

Hence required area is $\frac{{{\pi ^3}}}{4} - 2 = \frac{{{\pi ^3} - 8}}{4}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free