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• Notes

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
$x=4$

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

$(0,0)(1,1)(2,1.4)(3,1.7)(4,2)$

$y : 0 \rightarrow 2$

$x=4 \longrightarrow x=-1$

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

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

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

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

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

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

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

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