Notes: `y = x^3/6 + 1/(2x) , 1

Saturday, 30 August 2014

$\displaystyle{y}=\frac{{{x}}^{{3}}}{{6}}+\frac{{1}}{{{2}{x}}},{1}$

Area of the surface obtained by revolving the curve $\displaystyle{y}={f{{\left({x}\right)}}}$ about $\displaystyle{x}$ -axis between $\displaystyle{a}\leq{x}\leq{b}$ is given by

$\displaystyle{S}_{{x}}={2}\pi{\int_{{a}}^{{b}}}{y}\sqrt{{{1}+{y}{'}^{{2}}}}{\left.{d}{x}\right.}$

Let us therefore, first find $\displaystyle{y}'.$

$\displaystyle{y}'=\frac{{{x}}^{{2}}}{{2}}-\frac{{1}}{{{2}{{x}}^{{2}}}}=\frac{{{{x}}^{{4}}-{1}}}{{{2}{{x}}^{{2}}}}$

$\displaystyle{y}{'}^{{2}}=\frac{{{{x}}^{{8}}-{2}{{x}}^{{4}}+{1}}}{{{4}{{x}}^{{4}}}}$

We can now calculate the surface area.

$\displaystyle{S}_{{x}}={2}\pi{\int_{{1}}^{{2}}}{\left(\frac{{{x}}^{{3}}}{{6}}+\frac{{1}}{{{2}{x}}}\right)}\sqrt{{{1}+\frac{{{{x}}^{{8}}-{2}{{x}}^{{4}}+{1}}}{{{4}{{x}}^{{4}}}}}}{\left.{d}{x}\right.}=$

$\displaystyle{2}\pi{\int_{{1}}^{{2}}}{\left(\frac{{{x}}^{{3}}}{{6}}+\frac{{1}}{{{2}{x}}}\right)}\sqrt{{\frac{{{{x}}^{{8}}+{2}{{x}}^{{4}}+{1}}}{{{4}{{x}}^{{4}}}}}}=$

$\displaystyle{2}\pi{\int_{{1}}^{{2}}}{\left(\frac{{{x}}^{{3}}}{{6}}+\frac{{1}}{{{2}{x}}}\right)}\sqrt{{{{\left(\frac{{{{x}}^{{4}}+{1}}}{{{2}{{x}}^{{2}}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}=$

$\displaystyle{2}\pi{\int_{{1}}^{{2}}}{\left(\frac{{{x}}^{{3}}}{{6}}+\frac{{1}}{{{2}{x}}}\right)}\frac{{{{x}}^{{4}}+{1}}}{{{2}{{x}}^{{2}}}}{\left.{d}{x}\right.}=$

Multiplying the terms under integral yields

$\displaystyle{2}\pi{\int_{{1}}^{{2}}}{\left(\frac{{{x}}^{{5}}}{{12}}+\frac{{x}}{{12}}+\frac{{x}}{{4}}+\frac{{1}}{{{4}{{x}}^{{3}}}}\right)}{\left.{d}{x}\right.}=$

$\displaystyle{2}\pi{\int_{{1}}^{{2}}}{\left(\frac{{{x}}^{{5}}}{{12}}+\frac{{x}}{{3}}+\frac{{1}}{{{4}{{x}}^{{3}}}}\right)}{\left.{d}{x}\right.}=$

$\displaystyle{2}\pi{\left(\frac{{{x}}^{{6}}}{{72}}+\frac{{{x}}^{{2}}}{{6}}-\frac{{1}}{{{8}{{x}}^{{2}}}}\right)}{{\mid}_{{1}}^{{2}}}=$

$\displaystyle{2}\pi{\left(\frac{{64}}{{72}}+\frac{{2}}{{3}}-\frac{{1}}{{32}}-\frac{{1}}{{72}}-\frac{{1}}{{6}}+\frac{{1}}{{8}}\right)}={2}\pi\cdot\frac{{47}}{{32}}=\frac{{{47}\pi}}{{16}}$

The area of surface generated by revolving the...