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Math  /  Calculus

QuestionWhich of the following integrals represents the volume of the solid obtained by rotating the region bounded by the curves y=(x−2)4y=(x-2)^{4} and 8x−y=168 x-y=16 about the line x=10x=10 ? A. π∫016([10−(18y+2)]−[10−(2+y4)])2dy\pi \int_{0}^{16}\left(\left[10-\left(\frac{1}{8} y+2\right)\right]-[10-(2+\sqrt[4]{y})]\right)^{2} d y B. π∫24([10−(18y+2)]−[10−(2+y4)])2dy\pi \int_{2}^{4}\left(\left[10-\left(\frac{1}{8} y+2\right)\right]-[10-(2+\sqrt[4]{y})]\right)^{2} d y C. π∫016([10−(18y+2)]2−[10−(2+y4)]2)dy\pi \int_{0}^{16}\left(\left[10-\left(\frac{1}{8} y+2\right)\right]^{2}-[10-(2+\sqrt[4]{y})]^{2}\right) d y D. π∫016([10−(18y+2)2]−[10−(2+y4)2])dy\pi \int_{0}^{16}\left(\left[10-\left(\frac{1}{8} y+2\right)^{2}\right]-\left[10-(2+\sqrt[4]{y})^{2}\right]\right) d y E. π∫24([10−(18y+2)2]−[10−(2+y4)2])dy\pi \int_{2}^{4}\left(\left[10-\left(\frac{1}{8} y+2\right)^{2}\right]-\left[10-(2+\sqrt[4]{y})^{2}\right]\right) d y F. π∫24([10−(18y+2)]2−[10−(2+y4)]2)dy\pi \int_{2}^{4}\left(\left[10-\left(\frac{1}{8} y+2\right)\right]^{2}-[10-(2+\sqrt[4]{y})]^{2}\right) d y

Studdy Solution
Match the derived integral to the given choices.
π∫016([10−(18y+2)]−[10−(2+y4)])2dy\pi \int_{0}^{16} \left(\left[10 - \left(\frac{1}{8} y + 2\right)\right] - \left[10 - (2 + \sqrt[4]{y})\right]\right)^2 dy
This matches option (A).
Solution: Option (A) is the correct integral. A\boxed{A}

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