Notes: `y = e^(-x), y = 1, x = 2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

Sunday, 2 November 2014

$\displaystyle{y}={{e}}^{{-{x}}},{y}={1},{x}={2}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line....

You need to find the volume of the solid obtained by rotating the region enclosed by the curves $\displaystyle{y}={{e}}^{{-{x}}},{y}={1},{x}={2}$ , about y = 2, using washer method:

$\displaystyle{V}=\pi\cdot{\int_{{a}}^{{b}}}{\left({{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.},{f{{\left({x}\right)}}}>{g{{\left({x}\right)}}}$

You need to find the endpoints by solving the equation:

$\displaystyle{{e}}^{{-{x}}}={1}\Rightarrow\frac{{1}}{{{{e}}^{{x}}}}={1}\Rightarrow{{e}}^{{x}}={1}\Rightarrow{{e}}^{{x}}={{e}}^{{0}}\Rightarrow{x}={0}$

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

You need to find the volume of the solid obtained by rotating the region enclosed by the curves $\displaystyle{y}={{e}}^{{-{x}}},{y}={1},{x}={2}$ , about y = 2, using washer method:

$\displaystyle{V}=\pi\cdot{\int_{{a}}^{{b}}}{\left({{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.},{f{{\left({x}\right)}}}>{g{{\left({x}\right)}}}$

You need to find the endpoints by solving the equation:

$\displaystyle{{e}}^{{-{x}}}={1}\Rightarrow\frac{{1}}{{{{e}}^{{x}}}}={1}\Rightarrow{{e}}^{{x}}={1}\Rightarrow{{e}}^{{x}}={{e}}^{{0}}\Rightarrow{x}={0}$

$\displaystyle{V}=\pi\cdot{\int_{{0}}^{{2}}}{\left({{\left({{e}}^{{-{x}}}-{2}\right)}}^{{2}}-{{\left({1}-{2}\right)}}^{{2}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{0}}^{{2}}}{\left({{e}}^{{-{2}{x}}}-{4}{{e}}^{{-{2}{x}}}+{4}-{1}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot{\int_{{0}}^{{2}}}{\left({{e}}^{{-{2}{x}}}-{4}{{e}}^{{-{2}{x}}}+{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}=\pi\cdot{\left(-\frac{{{{e}}^{{-{2}{x}}}}}{{2}}+{2}{{e}}^{{-{2}{x}}}+{3}{x}\right)}{{\mid}_{{0}}^{{2}}}$

$\displaystyle{V}=\pi\cdot{\left(-\frac{{{{e}}^{{-{4}}}}}{{2}}+{2}{{e}}^{{-{4}}}+{6}+\frac{{1}}{{2}}-{2}+{0}\right)}$

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

$\displaystyle{V}=\pi\cdot\frac{{-{1}+{4}+{9}{{e}}^{{4}}}}{{{2}{{e}}^{{4}}}}$

$\displaystyle{V}=\pi\cdot\frac{{{3}+{9}{{e}}^{{4}}}}{{{2}{{e}}^{{4}}}}$

Hence, evaluating the volume of the solid obtained by rotating the region enclosed by the curves $\displaystyle{y}={{e}}^{{-{x}}},{y}={1},{x}={2}$ , about y = 2, using washer method, yields $\displaystyle{V}=\pi\cdot\frac{{{3}+{9}{{e}}^{{4}}}}{{{2}{{e}}^{{4}}}}.$