`x = y^2 + 1, x = 2` Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves...

Saturday, April 10, 2010

$\displaystyle{x}={{y}}^{{2}}+{1},{x}={2}$ Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves...

The shell has the radius $\displaystyle-{2}-{y}$ , the cricumference is $\displaystyle{2}\pi\cdot{\left(-{2}-{y}\right)}$ and the height is $\displaystyle{2}-{x}$ , hence, the volume can be evaluated, using the method of cylindrical shells, such that:

$\displaystyle{V}={2}\pi\cdot{\int_{{{y}_{{1}}}}^{{{y}_{{2}}}}}{\left(-{2}-{y}\right)}\cdot{\left({2}-{x}\right)}{\left.{d}{y}\right.}$

You need to evaluate the endpoints from equation $\displaystyle{{y}}^{{2}}+{1}={2}\Rightarrow{{y}}^{{2}}={1}\Rightarrow{y}_{{{1},{2}}}=\pm{1}$

$\displaystyle{V}={2}\pi\cdot{\int_{{-{1}}}^{{1}}}{\left(-{2}-{y}\right)}\cdot{\left({2}-{{y}}^{{2}}-{1}\right)}{\left.{d}{y}\right.}$

$\displaystyle{V}={2}\pi\cdot{\int_{{-{1}}}^{{1}}}{\left(-{2}-{y}\right)}\cdot{\left({1}-{{y}}^{{2}}\right)}{\left.{d}{y}\right.}$

$\displaystyle{V}={2}\pi\cdot{\int_{{-{1}}}^{{1}}}{\left(-{2}+{2}{{y}}^{{2}}-{y}+{{y}}^{{3}}\right)}{\left.{d}{y}\right.}$

$\displaystyle{V}={2}\pi\cdot{\left({\int_{{-{1}}}^{{1}}}{\left(-{2}{\left.{d}{y}\right.}\right)}+{\int_{{-{1}}}^{{1}}}{2}{{y}}^{{2}}{\left.{d}{y}\right.}-{\int_{{-{1}}}^{{1}}}{y}{\left.{d}{y}\right.}+{\int_{{-{1}}}^{{1}}}{{y}}^{{3}}{\left.{d}{y}\right.}\right)}$

Using the formula $\displaystyle\int{{x}}^{{n}}{\left.{d}{x}\right.}=\frac{{{{x}}^{{{n}+{1}}}}}{{{n}+{1}}}$ yields:

$\displaystyle{V}={2}\pi\cdot{\left(-{2}{y}+{2}\frac{{{y}}^{{3}}}{{3}}-\frac{{{y}}^{{2}}}{{2}}+\frac{{{y}}^{{4}}}{{4}}\right)}{{\mid}_{{-{1}}}^{{1}}}$

$\displaystyle{V}={2}\pi\cdot{\left(-{2}+\frac{{2}}{{3}}-\frac{{1}}{{2}}+\frac{{1}}{{4}}+{2}+\frac{{2}}{{3}}+\frac{{1}}{{2}}-\frac{{1}}{{4}}\right)}{{\mid}_{{-{1}}}^{{1}}}$