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###### ${selected_topic_name} • Notes • Comments & Questions The region R enclosed by the curves y=x and $y=x^{2}$ is rotated about the x-axis. Find the volume of the resulting solid. y=x $\begin{array}{l|l}{x} & {y} \\ \hline 0 & {0} \\ {1} & {1} \\ {2} & {2} \\ {3} & {3}\end{array}$ $\begin{array}{l}{(0,0)} \\ {(1,1)} \\ {(2, 2)} \\ {(3, 3)}\end{array}$ $y={x^{2}}$ $\begin{array}{l|l}{x} & {y} \\ \hline {0} & {0} \\ {1} & {1} \\ {2} & {4} \\ {3} & {9}\end{array}$$\$

$\begin{array}{l}{(0,0)} \\ {(1,1)} \\ {(2,4)} \\ {(3,9)}\end{array}$

inter section points
$\longrightarrow x=x^{2} \longrightarrow x^{2}-x=0 \rightarrow x(x-1)=0$

إما  $x=0$

أو   $x-1=0 \rightarrow x=1$

$y=0 \quad y=1$

$(0,0) \quad (1,1)$

(1) washer method:
$d x : 0 \rightarrow 1 \quad { r_{out} } : x-0=x \quad {r_{in}} : x^{2}-0=x^{2}$

volume:
$=\pi \int_{0}^{1}\left[{r_{out}}^{2}-{r_ { i n}}^{2}\right] d x$

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

$=\pi \left[\frac{x^{3}}{3}-\frac{x^{5}}{5}\right]_{0}^{1}$

$=\pi\left[\left(\frac{(1)^{3}}{3}-\frac{(1)^{5}}{5}\right)-(0)\right) ]=\pi\left(\frac{1}{3}-\frac{1}{5}\right)$

$=\pi\left[\frac{5-3}{15}\right]=\frac{2 \pi}{15}$

(2) c s M :

$d y : 0 \rightarrow \quad x$

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

$x=y-0=y$

$f(x)=x_{right} - x_{left} = \sqrt y - y$

volume:
$=2 \pi \int_{1}^{1} x f(x) d y$

volume:
$=2 \pi \int_{0}^{1} y(\sqrt{y}-y) d y$

$=2 \pi \int_{0}^{1}\left(y \cdot \sqrt{y}-y^{2}\right) d y=2 \pi \int_{0}^{1} y \cdot y^{\frac{1}{2}}-y^{2} d y$

$=2 \pi \int_{0}^{1}\left(y^{\frac{3}{2}}-y^{2}\right) d y$

$=2 \pi\left[\frac{2}{5} y^{\frac{5}{2}}-\frac{y^{3}}{3}\right]_{0}^{1}=2 \pi\left[\left(\frac{2}{5}(1)^{\frac{5}{2}}-\frac{(1)^{3}}{3}\right)-(0)\right]$

$=2 \pi\left[\frac{6-5}{15}\right]=\frac{2 \pi}{15}$