Notes: `y = 2 - (1/2)x, y = 0, x = 1, x = 2` Find the volume of the solid obtained by rotating the region bounded by the given curves about the...

Tuesday, 3 September 2013

$\displaystyle{y}={2}-{\left(\frac{{1}}{{2}}\right)}{x},{y}={0},{x}={1},{x}={2}$ Find the volume of the solid obtained by rotating the region bounded by the given curves about the...

The volume of the solid obtained by rotating the region bounded by the curves $\displaystyle{y}={2}-\frac{{x}}{{2}},{y}={0},{x}={1},{x}={2}$ , about x axis, can be evaluated using the washer method, such that:

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

Since the problem provides you the endpoints x=1,x=2, you may find the volume such that:

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

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

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

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

Since the problem provides you the endpoints x=1,x=2, you may find the volume such that:

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

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

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

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

$\displaystyle{V}=\pi\cdot{\left({x}-{{x}}^{{2}}+\frac{{{x}}^{{3}}}{{12}}\right)}{{\mid}_{{1}}^{{2}}}$

$\displaystyle{V}=\pi\cdot{\left(-{2}+{{2}}^{{2}}-\frac{{{2}}^{{3}}}{{12}}-{1}+{{1}}^{{2}}+\frac{{{1}}^{{3}}}{{12}}\right)}$

$\displaystyle{V}=\pi\cdot{\left({2}-\frac{{8}}{{12}}+\frac{{1}}{{12}}\right)}$

$\displaystyle{V}=\frac{{{17}\pi}}{{12}}$

Hence, evaluating the volume of the solid obtained by rotating the region bounded by the curves $\displaystyle{y}={2}-\frac{{x}}{{2}},{y}={0},{x}={1},{x}={2}$ , about x axis , using the washer method, yields $\displaystyle{V}=\frac{{{17}\pi}}{{12}}.$