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Find the volume of the solid obtained by rotating the region R about the axis (\(x=-1\)) where the region $R$ is specified between \(y=\sqrt{x} \quad y=0\) and \(x=4\)

\(\begin{equation} \begin{array}{l|l}{x}&{y} \\ {0} & {0} \\ {1} & {1} \\ {2} & {1.4} \\ {3} & {1.7} \\ {4} & {2}\end{array} \end{equation}\)


\(x : 0 \longrightarrow 4\)
 


\(x=x-(-1)=x+1\)
 


\(f(x)=y_{u p}-y_{d o w{n}}=\sqrt{x}-0=\sqrt{x} \)

volume:
\(v=2 \pi \int_{0}^{4} x f(x) d x \)


\(=2 \pi \int_{0}^{4}({x+1 )(\sqrt{x})} d x\)
 


\(=2 \pi \int_{0}^{4}(x \sqrt{x}+\sqrt{x}) d x=2 \pi \int_{0}^{4}\left(x x^{\frac{1}{2}}+x^{\frac{1}{2}}\right) d x \)


\(=2 \pi \int_{0}^{4}\left(x^{\frac{3}{2}}+x^{\frac{1}{2}}\right) d x=2 \pi\left[\frac{2}{5} x^{\frac{5}{2}}+\frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{4}\)
 


\(=2 \pi\left[\frac{2}{5}(4)^{\frac{5}{2}}+\frac{2}{3}(4)^{\frac{3}{2}}\right]-[0] \)


\(=2 \pi\left[\frac{2}{5}(4)^{\frac{5}{2}}+\frac{2}{3}(4)^{\frac{3}{2}}\right]=\frac{544 \pi}{15} \)

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