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Question

Mathematics Question on Area between Two Curves

The area in the first quadrant between x2+y2=π2x^2 + y^2 = \pi^2 and y=sin⁡ xy = \sin \, x is

A

π3−84\frac{\pi^3 - 8 }{4}

B

π34\frac{\pi^3}{4}

C

π3−164\frac{\pi^3 - 16 }{4}

D

π3−82\frac{\pi^3 - 8 }{2}

Answer

π3−84\frac{\pi^3 - 8 }{4}

Explanation

Solution

x2+y2=π2x^2 + y^2 = \pi^2 is a circle of radius π\pi and centre at origin.

Required area
= Area of circle (1st quadrant ) - ∫0π sin⁡ x dx\int\limits^{\pi}_0 \, \sin \, x \, dx
=π.π24−[−cos⁡x]0π= \frac{\pi.\pi^{2}}{4} - \left[-\cos x \right]^{\pi}_{0}
=π24+(cos⁡π−cos⁡0)= \frac{\pi^{2}}{4}+ \left(\cos\pi - \cos0\right)
=π34+(−1−1)=π34−2= \frac{\pi^{3}}{4} + \left(-1-1\right)= \frac{\pi^{3}}{4} - 2
=π3−84= \frac{\pi^{3} -8}{ 4}