Notes: `y = sqrt(x), y = 0, x = 1` Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given...

Friday, 17 March 2017

$\displaystyle{y}=\sqrt{{{x}}},{y}={0},{x}={1}$ Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given...

With the method of cylindrical shells we sum up the volumes of thin cylinders.

The volume of a cylinder is

$\displaystyle{2}\pi\cdot{r}\cdot{h}\cdot{d}{r},$

where $\displaystyle{h}$ is the height (the value of a function), $\displaystyle{r}$ is the radius of a cylinder (the distance from the axis of rotation to the argument) and $\displaystyle{d}{r}$ is the thickness.

Here this is $\displaystyle{2}\pi\cdot{\left({x}+{1}\right)}\cdot\sqrt{{{x}}}{\left.{d}{x}\right.}$ and the volume is

$\displaystyle{2}\pi{\int_{{0}}^{{1}}}{\left({x}+{1}\right)}\sqrt{{{x}}}{\left.{d}{x}\right.}={2}\pi{\int_{{0}}^{{1}}}{\left({{x}}^{{\frac{{3}}{{2}}}}+{{x}}^{{\frac{{1}}{{2}}}}\right)}{\left.{d}{x}\right.}=$

$\displaystyle={2}\pi{{\left(\frac{{2}}{{{5}}}{{x}}^{{\frac{{5}}{{2}}}}+\frac{{2}}{{{3}}}{{x}}^{{\frac{{3}}{{2}}}}\right)}_{{0}}^{{1}}}={2}\pi{\left(\frac{{2}}{{5}}+\frac{{2}}{{3}}\right)}=\frac{{32}}{{{15}}}\pi.$