Definite Integration Question 297
Question: The area in the first quadrant between $ x^{2}+y^{2}={{\pi }^{2}} $ and $ y=\sin x $ is
[MP PET 1997]
Options:
A) $ \frac{({{\pi }^{3}}-8)}{4} $
B) $ \frac{{{\pi }^{3}}}{4} $
C) $ \frac{({{\pi }^{3}}-16)}{4} $
D) $ \frac{({{\pi }^{3}}-8)}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
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} $ .
BETA
