The area in the first quadrant between x^2 + | vedclass

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(A) The equation of the circle is $x^2 + y^2 = \pi^2$,which represents a circle with radius $r = \pi$. In the first quadrant,the area of this circle i

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 = \pi^2$,which represents a circle with radius $r = \pi$. In the first quadrant,the area of this circle is $\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (\pi^2) = \frac{\pi^3}{4}$. The area bounded by the curve $y = \sin x$ and the $x$-axis in the first quadrant (from $x = 0$ to $x = \pi$) is given by $\int_{0}^{\pi} \sin x \, dx$. Evaluating this integral: $\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$. The required area is the area of the quarter circle minus the area under the sine curve: $\frac{\pi^3}{4} - 2 = \frac{\pi^3 - 8}{4}$.

The area in the first quadrant between $x^2 + y^2 = \pi^2$ and $y = \sin x$ is

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