Let be the angle at which the curves and intersect in the first quadrant

MATHEMATICS · Single correct · AP · Calculus BC · Parametric, Polar, and Vector-Valued Functions · Parametric, Polar, and Vector-Valued Functions

Problem

Let α=π4\alpha=\dfrac{\pi}{4} be the angle at which the curves r=2sin⁡θr=2\sin\theta and r=2cos⁡θr=2\cos\theta intersect in the first quadrant. Which of the following is an exact expression for the area of the region bounded by the two curves in the first quadrant? (a) 12∫0π/2((2sin⁡θ)2−(2cos⁡θ)2) dθ\tfrac12\displaystyle\int_{0}^{\pi/2}\big((2\sin\theta)^2-(2\cos\theta)^2\big)\,d\theta (b) 12[∫0α((2sin⁡θ)2−(2cos⁡θ)2) dθ+∫απ/2((2cos⁡θ)2−(2sin⁡θ)2) dθ]\tfrac12\Bigg[\displaystyle\int_{0}^{\alpha}\big((2\sin\theta)^2-(2\cos\theta)^2\big)\,d\theta+\int_{\alpha}^{\pi/2}\big((2\cos\theta)^2-(2\sin\theta)^2\big)\,d\theta\Bigg] (c) 12[∫0α((2cos⁡θ)2−(2sin⁡θ)2) dθ+∫απ/2((2sin⁡θ)2−(2cos⁡θ)2) dθ]\tfrac12\Bigg[\displaystyle\int_{0}^{\alpha}\big((2\cos\theta)^2-(2\sin\theta)^2\big)\,d\theta+\int_{\alpha}^{\pi/2}\big((2\sin\theta)^2-(2\cos\theta)^2\big)\,d\theta\Bigg] (d) 12[∫0α((2cos⁡θ)2−(2sin⁡θ)2) dθ−∫απ/2((2cos⁡θ)2−(2sin⁡θ)2) dθ]\tfrac12\Bigg[\displaystyle\int_{0}^{\alpha}\big((2\cos\theta)^2-(2\sin\theta)^2\big)\,d\theta-\int_{\alpha}^{\pi/2}\big((2\cos\theta)^2-(2\sin\theta)^2\big)\,d\theta\Bigg]

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