How-can-I-calculate-the-volume-of-a-region-bounded-by-y-x-2-3-x-1-and-x-2-rotating-about-the-y-7-using-the-shell-method-

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Question Number 131364 by EDWIN88 last updated on 04/Feb/21

$$$$$$\mathrm{How}\mathrm{can}\mathrm{I}\mathrm{calculate}\mathrm{the}\mathrm{volume}$$$$\mathrm{of}\mathrm{a}\mathrm{region}\mathrm{bounded}\mathrm{by}\mathrm{y}={\mathrm{x}}^2+3;\mathrm{x}=1$$$$\mathrm{and}\mathrm{x}=2\mathrm{rotating}\mathrm{about}\mathrm{the}\mathrm{y}=7\mathrm{using}$$$$\mathrm{the}\mathrm{shell}\mathrm{method}.$$

Answered by bramlexs22 last updated on 04/Feb/21

$$V=2\pi \underset{4}{\stackrel{7}{\int }}(7-y)(\sqrt{y-3}-2)dy$$$$let\sqrt{y-3}=q\to \overline{)\begin{array}{c}y=7\to q=2\\ y=4\to q=1\end{array}}anddy=2qdq$$$$V=2\pi \underset{1}{\stackrel{2}{\int }}(7-(q^2+3))(q-2)\left(2qdq\right)$$$$V=4\pi \underset{1}{\stackrel{2}{\int }}(4-q^2)(q^2-2q)dq$$$$V=4\pi \underset{1}{\stackrel{2}{\int }}(-8q+4q^2+2q^3-q^4)dq$$$$V=4\pi {[-4q^2+\frac{4q^3}{3}+\frac{q^4}{2}-\frac{q^5}{5}]}_1^2$$$$V=4\pi [-4\left(3\right)+\frac{4\left(7\right)}{3}+\frac{15}{2}-\frac{31}{5}]$$$$V=4\pi [\frac{-180+420-93}{15}+\frac{15}{2}]=4\pi [\frac{147}{15}+\frac{15}{2}]$$$$V=4\pi \left[\frac{298+225}{30}\right]=\frac{1046\pi }{15}$$$$$$

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