Let be the solid generated when the region bounded by and on the interval is revolved…

MATHEMATICS · Single correct · AP · Calculus AB · Applications of Integration · Applications of Integration

Problem

Let SS be the solid generated when the region bounded by y=x+2y = x + 2 and y=(x−1)2y = (x - 1)^{2} on the interval −1≤x≤3-1 \le x \le 3 is revolved about the line y=−1y = -1. Which of the following integrals represents the volume of SS? (a) π∫−13((x+3)2−((x−1)2+1)2) dx\displaystyle \pi\int_{-1}^{3}\big((x+3)^{2}-((x-1)^{2}+1)^{2}\big)\,dx (b) π[∫−13−132((x−1)2)2−(x+3)2 dx+∫3−1323((x+3)2−((x−1)2)2) dx]\displaystyle \pi\left[\int_{-1}^{\frac{3-\sqrt{13}}{2}}\big((x-1)^{2}\big)^{2}-(x+3)^{2}\,dx+\int_{\frac{3-\sqrt{13}}{2}}^{3}\big((x+3)^{2}-((x-1)^{2})^{2}\big)\,dx\right] (c) π[∫−13−132((x−1)2+1)2−(x+3)2 dx+∫3−1323((x+3)2−((x−1)2+1)2) dx]\displaystyle \pi\left[\int_{-1}^{\frac{3-\sqrt{13}}{2}}\big((x-1)^{2}+1\big)^{2}-(x+3)^{2}\,dx+\int_{\frac{3-\sqrt{13}}{2}}^{3}\big((x+3)^{2}-((x-1)^{2}+1)^{2}\big)\,dx\right] (d) π[∫−13−132(x+3)2−((x−1)2+1)2 dx+∫3−1323((x−1)2+1)2−(x+3)2 dx]\displaystyle \pi\left[\int_{-1}^{\frac{3-\sqrt{13}}{2}}(x+3)^{2}-((x-1)^{2}+1)^{2}\,dx+\int_{\frac{3-\sqrt{13}}{2}}^{3}\big((x-1)^{2}+1\big)^{2}-(x+3)^{2}\,dx\right]

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