What is the area of the region enclosed in the first quadrant by x^{2}+y^{2}=\pi^{2}, y=\sin x and x=0?
- A. \frac{\pi^{3}}{4}-1
- B. \frac{\pi^{3}}{4}-2 ✓
- C. \frac{\pi^{3}}{2}-1
- D. \frac{\pi^{2}}{4}-2
Correct Answer: B. \frac{\pi^{3}}{4}-2
Explanation
The region in the first quadrant lies between the circle y = \sqrt{\pi^2-x^2} and the curve y = \sin x. The circle intersects the x-axis at x = \pi, which matches the root of \sin x in the first quadrant. The area is \int_0^\pi (\sqrt{\pi^2-x^2} - \sin x) \,dx. The integral of the circle is the area of a quarter circle of radius \pi: \frac{1}{4}\pi(\pi)^2 = \frac{\pi^3}{4}. The integral of \sin x from 0 to \pi is 2. Subtracting them gives \frac{\pi^3}{4} - 2.
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