`y = xe^(-x), y = 0, x = 2` (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

Tuesday, June 23, 2009

$\displaystyle{y}={x}{{e}}^{{-{x}}},{y}={0},{x}={2}$ (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

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

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

You need to find the next endpoint, using the equation $\displaystyle{x}\cdot{{e}}^{{-{x}}}={0}\Rightarrow{x}={0}$

$\displaystyle{V}={2}\pi\cdot{\int_{{0}}^{{2}}}{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}$

You need to use integration by parts to evaluate the volume, such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={{x}}^{{2}}\Rightarrow{d}{u}={2}{x}{\left.{d}{x}\right.}$

$\displaystyle{d}{v}={{e}}^{{-{x}}}\Rightarrow{v}=-{{e}}^{{-{x}}}$

$\displaystyle{\int_{{0}}^{{2}}}{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{{\mid}_{{0}}^{{2}}}+{2}{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}$

You need to use integration by parts to evaluate the integral $\displaystyle{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}.$

$\displaystyle{u}={x}\Rightarrow{d}{u}={\left.{d}{x}\right.}$

$\displaystyle{d}{v}={{e}}^{{-{x}}}\Rightarrow{v}=-{{e}}^{{-{x}}}$

$\displaystyle{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-{x}\cdot{{e}}^{{-{x}}}{{\mid}_{{0}}^{{2}}}+{\int_{{0}}^{{2}}}{{e}}^{{-{x}}}{\left.{d}{x}\right.}$

$\displaystyle{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-{x}\cdot{{e}}^{{-{x}}}{{\left|_{{0}}^{{2}}-{{e}}^{{-{x}}}\right|}_{{0}}^{{2}}}$

$\displaystyle{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-{2}\cdot{{e}}^{{-{2}}}-{{e}}^{{-{2}}}+{0}\cdot{{e}}^{{{0}}}+{{e}}^{{{0}}}$

$\displaystyle{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-\frac{{2}}{{{{e}}^{{2}}}}-\frac{{1}}{{{{e}}^{{2}}}}+{1}$

$\displaystyle{\int_{{0}}^{{2}}}{x}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-\frac{{3}}{{{{e}}^{{2}}}}+{1}$

$\displaystyle{\int_{{0}}^{{2}}}{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{{\mid}_{{0}}^{{2}}}+{2}{\left(-\frac{{3}}{{{{e}}^{{2}}}}+{1}\right)}$

$\displaystyle{\int_{{0}}^{{2}}}{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-{{2}}^{{2}}\cdot{{e}}^{{-{2}}}-\frac{{6}}{{{{e}}^{{2}}}}+{2}$

$\displaystyle{\int_{{0}}^{{2}}}{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-\frac{{4}}{{{{e}}^{{2}}}}-\frac{{6}}{{{{e}}^{{2}}}}+{2}$

$\displaystyle{\int_{{0}}^{{2}}}{{x}}^{{2}}\cdot{{e}}^{{-{x}}}{\left.{d}{x}\right.}=-\frac{{10}}{{{{e}}^{{2}}}}+{2}$

$\displaystyle{V}={2}\pi\cdot{\left(-\frac{{10}}{{{{e}}^{{2}}}}+{2}\right)}$

Hence, evaluating the volume, using the method of cylindrical shells, yields $\displaystyle{V}={2}\pi\cdot{\left(-\frac{{10}}{{{{e}}^{{2}}}}+{2}\right)}.$

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Tuesday, June 23, 2009

$\displaystyle{y}={x}{{e}}^{{-{x}}},{y}={0},{x}={2}$ (a) Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curve...

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