Question Number 131364 by EDWIN88 last updated on 04/Feb/21
$$ \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{I}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{volume}\: \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{3}\:;\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{x}=\mathrm{2}\:\mathrm{rotating}\:\mathrm{about}\:\mathrm{the}\:\mathrm{y}=\mathrm{7}\:\mathrm{using} \\ $$$$\mathrm{the}\:\mathrm{shell}\:\mathrm{method}. \\ $$
Answered by bramlexs22 last updated on 04/Feb/21
![V= 2π∫_4 ^7 (7−y)((√(y−3))−2) dy let (√(y−3)) = q → determinant (((y=7→q=2)),((y=4→q=1)))and dy=2q dq V=2π∫_1 ^2 (7−(q^2 +3))(q−2)(2q dq) V=4π∫_1 ^2 (4−q^2 )(q^2 −2q)dq V=4π∫_1 ^2 (−8q+4q^2 +2q^3 −q^4 )dq V=4π [−4q^2 +((4q^3 )/3)+(q^4 /2)−(q^5 /5) ]_1 ^2 V=4π [ −4(3)+((4(7))/3)+((15)/2)−((31)/5) ] V=4π [ ((−180+420−93)/(15))+((15)/2) ]=4π [ ((147)/(15))+((15)/2)] V=4π [ ((298+225)/(30)) ]= ((1046π)/(15))](https://www.tinkutara.com/question/Q131410.png)
$$\:{V}=\:\mathrm{2}\pi\underset{\mathrm{4}} {\overset{\mathrm{7}} {\int}}\:\left(\mathrm{7}−{y}\right)\left(\sqrt{{y}−\mathrm{3}}−\mathrm{2}\right)\:{dy}\: \\ $$$$\:{let}\:\sqrt{{y}−\mathrm{3}}\:=\:{q}\:\rightarrow\begin{array}{|c|c|}{{y}=\mathrm{7}\rightarrow{q}=\mathrm{2}}\\{{y}=\mathrm{4}\rightarrow{q}=\mathrm{1}}\\\hline\end{array}{and}\:{dy}=\mathrm{2}{q}\:{dq} \\ $$$${V}=\mathrm{2}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\left(\mathrm{7}−\left({q}^{\mathrm{2}} +\mathrm{3}\right)\right)\left({q}−\mathrm{2}\right)\left(\mathrm{2}{q}\:{dq}\right) \\ $$$${V}=\mathrm{4}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\left(\mathrm{4}−{q}^{\mathrm{2}} \right)\left({q}^{\mathrm{2}} −\mathrm{2}{q}\right){dq} \\ $$$${V}=\mathrm{4}\pi\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\left(−\mathrm{8}{q}+\mathrm{4}{q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} −{q}^{\mathrm{4}} \right){dq} \\ $$$${V}=\mathrm{4}\pi\:\left[−\mathrm{4}{q}^{\mathrm{2}} +\frac{\mathrm{4}{q}^{\mathrm{3}} }{\mathrm{3}}+\frac{{q}^{\mathrm{4}} }{\mathrm{2}}−\frac{{q}^{\mathrm{5}} }{\mathrm{5}}\:\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${V}=\mathrm{4}\pi\:\left[\:−\mathrm{4}\left(\mathrm{3}\right)+\frac{\mathrm{4}\left(\mathrm{7}\right)}{\mathrm{3}}+\frac{\mathrm{15}}{\mathrm{2}}−\frac{\mathrm{31}}{\mathrm{5}}\:\right] \\ $$$${V}=\mathrm{4}\pi\:\left[\:\frac{−\mathrm{180}+\mathrm{420}−\mathrm{93}}{\mathrm{15}}+\frac{\mathrm{15}}{\mathrm{2}}\:\right]=\mathrm{4}\pi\:\left[\:\frac{\mathrm{147}}{\mathrm{15}}+\frac{\mathrm{15}}{\mathrm{2}}\right] \\ $$$${V}=\mathrm{4}\pi\:\left[\:\frac{\mathrm{298}+\mathrm{225}}{\mathrm{30}}\:\right]=\:\frac{\mathrm{1046}\pi}{\mathrm{15}} \\ $$$$ \\ $$