Mathematics

Mathematics

Problem

Which definite integral gives the area of the region that lies inside r=3cos⁡θr=3\cos\theta and outside r=3sin⁡θr=3\sin\theta in the first quadrant? a) 12∫0π/4((3sin⁡θ)2−(3cos⁡θ)2) dθ\displaystyle \frac{1}{2}\int_{0}^{\pi/4}\bigl((3\sin\theta)^2-(3\cos\theta)^2\bigr)\,d\theta b) ∫0π/4((3cos⁡θ)2−(3sin⁡θ)2) dθ\displaystyle \int_{0}^{\pi/4}\bigl((3\cos\theta)^2-(3\sin\theta)^2\bigr)\,d\theta c) 12∫0π/4((3cos⁡θ)2−(3sin⁡θ)2) dθ\displaystyle \frac{1}{2}\int_{0}^{\pi/4}\bigl((3\cos\theta)^2-(3\sin\theta)^2\bigr)\,d\theta d) 12∫0π/2((3cos⁡θ)2−(3sin⁡θ)2) dθ\displaystyle \frac{1}{2}\int_{0}^{\pi/2}\bigl((3\cos\theta)^2-(3\sin\theta)^2\bigr)\,d\theta

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