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Question: The area in the first quadrant between x<sup>2</sup> + y<sup>2</sup> = p<sup>2</sup> and y = sin x i...

The area in the first quadrant between x2 + y2 = p2 and y = sin x is -

A

(π3−8)4\frac { \left( \pi ^ { 3 } - 8 \right) } { 4 }

B

π34\frac { \pi ^ { 3 } } { 4 }

C

(π3−16)4\frac { \left( \pi ^ { 3 } - 16 \right) } { 4 }

D

(π3−8)2\frac { \left( \pi ^ { 3 } - 8 \right) } { 2 }

Answer

(π3−8)4\frac { \left( \pi ^ { 3 } - 8 \right) } { 4 }

Explanation

Solution

Area = ∫0π(y2−y1)dx\int _ { 0 } ^ { \pi } \left( y _ { 2 } - y _ { 1 } \right) d x

= ∫0π{π2−x2−sin⁡x}dx\int _ { 0 } ^ { \pi } \left\{ \sqrt { \pi ^ { 2 } - x ^ { 2 } } - \sin x \right\} d x

= dx – 2