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(D) The equation $x^2 + y^2 = \pi^2$ represents a circle centered at the origin with radius $r = \pi$.The area of the circle in the first quadrant is

Vedclass · Accepted Answer

$\frac{\pi^3-8}{4}$

Vedclass · Answer

(D) The equation $x^2 + y^2 = \pi^2$ represents a circle centered at the origin with radius $r = \pi$.The area of the circle in the first quadrant is given by $\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (\pi)^2 = \frac{\pi^3}{4}$ sq units.The curve $y = \sin x$ intersects the $x$-axis at $x = 0$ and $x = \pi$ in the first quadrant.The area under the curve $y = \sin x$ from $x = 0$ to $x = \pi$ is $\int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$ sq units.The required area between the curves in the first quadrant is the area of the quarter circle minus the area under the sine curve.Required Area $= \frac{\pi^3}{4} - 2 = \frac{\pi^3 - 8}{4}$ sq units.

Area lying in the first quadrant between the curves $x^2 + y^2 = \pi^2$ and $y = \sin x$ is equal to :-

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