• عربي

Need Help?

Subscribe to Calculus B

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

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

$x : 0 \longrightarrow 4$

$x=x-(-1)=x+1$

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

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

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

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

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

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

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