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Find the volume of the solid obtained by rotating the region about the axis
\((x=-1)\)
where the region R is specified between
\(y=\sqrt{x} \quad y=0\)

and 
\(x=4\)

 

\(\begin{equation} \begin{array}{l|l}{x} & {y} \\ {0} & {0} \\ {1} & {1} \\ {2} & {1 \cdot 4} \\ {3} & {1 \cdot 7} \\ {4} & {2}\end{array} \end{equation}\)


\((0,0)(1,1)(2,1.4)(3,1.7)(4,2)\)
 


\(y : 0 \rightarrow 2\)
 


\(x=4 \longrightarrow x=-1 \)


\(y=\sqrt{x} \rightarrow y^{2}=(\sqrt{x})^{2} \rightarrow y^{2}=x \rightarrow x = y^{2} \)


\(r_{i n}=y^{2}-(-1)=y^{2}+1 \)

volume: 
\(v=\pi \int_{0}^{2}\left[{v_{out}}^{2}-{r_{ i n}}^{2}\right] d y \)


\(=\pi \int_{0}^{2}\left[(5)^{2}-\left(y^{2}+1\right)^{2}\right] d y=\pi \int_{0}^{2}\left[25-\left(y^{4}+2 y^{2}+1\right)\right] d y \)


\(=\pi \int_{0}^{2}\left[25-y^{4}-2 y^{2}-1\right] d y \)


\(=\pi \int_{0}^{2}\left[24-y^{4}-2 y^{2}\right] d y=\pi\left[24 y-\frac{y^{5}}{5}-\frac{2}{3} y\right]_{0}^{2} \)


\(=\pi\left[\left(24(2)-\frac{(2)^{5}}{5}-\frac{2}{3}(2)^{3}\right)-\left(0-0-0\right)\right] \)


\(=\pi\left[48-\frac{32}{5}-\frac{16}{3}\right]=\frac{544\pi}{15} \)

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