Notes: `y = 4x - x^2, y = 3` Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves...

Wednesday, 13 September 2017

$\displaystyle{y}={4}{x}-{{x}}^{{2}},{y}={3}$ Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves...

$\displaystyle{y}={4}{x}-{{x}}^{{2}},{y}={3}$

The point of intersection of the curves will be ,

$\displaystyle{4}{x}-{{x}}^{{2}}={3}$

$\displaystyle{4}{x}-{{x}}^{{2}}-{3}={0}$

$\displaystyle-{{x}}^{{2}}+{4}{x}-{3}={0}$

$\displaystyle-{\left({{x}}^{{2}}-{4}{x}+{3}\right)}={0}$

$\displaystyle{{x}}^{{2}}-{4}{x}+{3}={0}$

factorizing the above equation,

$\displaystyle{\left({x}-{3}\right)}{\left({x}-{1}\right)}={0}$

x=3 , x=1

The shell has radius (x-1) , circumference $\displaystyle{2}\pi{\left({x}-{1}\right)}$ and height $\displaystyle{\left({4}{x}-{{x}}^{{2}}\right)}-{3}$

Volume generated by rotating the region bounded by the given curves about x=1 (V)= $\displaystyle{\int_{{1}}^{{3}}}{\left({2}\pi\right)}{\left({x}-{1}\right)}{\left({4}{x}-{{x}}^{{2}}-{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={\int_{{1}}^{{3}}}{\left({2}\pi\right)}{\left({4}{{x}}^{{2}}-{{x}}^{{3}}-{3}{x}-{4}{x}+{{x}}^{{2}}+{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={\left({2}\pi\right)}{\int_{{1}}^{{3}}}{\left(-{{x}}^{{3}}+{5}{{x}}^{{2}}-{7}{x}+{3}\right)}{\left.{d}{x}\right.}$

$\displaystyle{V}={2}\pi{{\left[-\frac{{{x}}^{{4}}}{{4}}+{5}\frac{{{x}}^{{3}}}{{3}}-{7}\frac{{{x}}^{{2}}}{{2}}+{3}{x}\right]}_{{1}}^{{3}}}$

$\displaystyle{V}={2}\pi{\left[-\frac{{{3}}^{{4}}}{{4}}+\frac{{5}}{{3}}\cdot{{3}}^{{3}}-\frac{{7}}{{2}}\cdot{{3}}^{{2}}+{3}\cdot{3}\right]}-{2}\pi{\left[-\frac{{{1}}^{{4}}}{{4}}+\frac{{5}}{{3}}\cdot{{1}}^{{3}}-\frac{{7}}{{2}}\cdot{{1}}^{{2}}+{3}\cdot{1}\right]}$

$\displaystyle{V}={2}\pi{\left({\left(-\frac{{81}}{{4}}+{45}-\frac{{63}}{{2}}+{9}\right)}-{\left(-\frac{{1}}{{4}}+\frac{{5}}{{3}}-\frac{{7}}{{2}}+{3}\right)}\right)}$

$\displaystyle{V}={2}\pi{\left(\frac{{9}}{{4}}-\frac{{11}}{{12}}\right)}$

$\displaystyle{V}={2}\pi{\left(\frac{{16}}{{12}}\right)}$

$\displaystyle{V}=\frac{{{8}\pi}}{{3}}$