The region bounded by and for is revolved about the vertical line

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

Problem

The region bounded by x=y2x=y^{2} and x=y+2x=y+2 for −1≤y≤2-1\le y\le 2 is revolved about the vertical line x=1x=1. Which of the following integrals equals the volume of the solid? (a) π∫−12((y+2)2−y4) dy\displaystyle \pi\int_{-1}^{2}\bigl((y+2)^{2}-y^{4}\bigr)\,dy (b) π∫−12((y+1)2−(y2−1)2) dy\displaystyle \pi\int_{-1}^{2}\bigl((y+1)^{2}-(y^{2}-1)^{2}\bigr)\,dy (c) π[∫−11((y+1)2−(1−y2)2) dy+∫12(y+1)2 dy]\displaystyle \pi\left[\int_{-1}^{1}\bigl((y+1)^{2}-(1-y^{2})^{2}\bigr)\,dy+\int_{1}^{2}(y+1)^{2}\,dy\right] (d) π∫−12((y+1)2−(1−y2)) dy\displaystyle \pi\int_{-1}^{2}\bigl((y+1)^{2}-(1-y^{2})\bigr)\,dy

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