Question:
The area in the first quadrant between $x^2 + y^2 = \pi^2$ and $y = \sin \, x$ is
The area in the first quadrant between $x^2 + y^2 = \pi^2$ and $y = \sin \, x$ is
Updated On: Jun 18, 2022
- $\frac{\pi^3 - 8 }{4}$
- $\frac{\pi^3}{4}$
- $\frac{\pi^3 - 16 }{4}$
- $\frac{\pi^3 - 8 }{2}$
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The Correct Option is A
Solution and Explanation
$x^2 + y^2 = \pi^2$ is a circle of radius $\pi $ and centre at origin.
Required area
= Area of circle (1st quadrant ) - $\int\limits^{\pi}_0 \, \sin \, x \, dx $
$= \frac{\pi.\pi^{2}}{4} - \left[-\cos x \right]^{\pi}_{0}$
$= \frac{\pi^{2}}{4}+ \left(\cos\pi - \cos0\right)$
$ = \frac{\pi^{3}}{4} + \left(-1-1\right)= \frac{\pi^{3}}{4} - 2$
$= \frac{\pi^{3} -8}{ 4}$
Required area
= Area of circle (1st quadrant ) - $\int\limits^{\pi}_0 \, \sin \, x \, dx $
$= \frac{\pi.\pi^{2}}{4} - \left[-\cos x \right]^{\pi}_{0}$
$= \frac{\pi^{2}}{4}+ \left(\cos\pi - \cos0\right)$
$ = \frac{\pi^{3}}{4} + \left(-1-1\right)= \frac{\pi^{3}}{4} - 2$
$= \frac{\pi^{3} -8}{ 4}$
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