Damon Vazquez

2022-09-07

Have I incorrectly set up the integral for this center of gravity/centroid problem?
Consider the bounded region R between the graphs of $g\left(x\right)=3x$ and g(x)=3x. Sketch R, which appropriate labels. Then set up all the integrals required do not evaluate.
My solution:
Definitions:
$A={\int }_{a}^{b}\left(f\left(x\right)-g\left(x\right)\right)dx$
${M}_{x}=y-\text{bar}=\frac{1}{A}{\int }_{a}^{b}\frac{1}{2}\left(f\left(x{\right)}^{2}-g\left(x{\right)}^{2}\right)dx$
${M}_{y}=x-\text{bar}=\frac{1}{A}{\int }_{a}^{b}x\left(f\left(x\right)-g\left(x\right)\right)dx$
….
$A={\int }_{0}^{3}\left({x}^{2}-3x\right)dx=\frac{9}{2}$
${M}_{x}=y-\text{bar}=\frac{2}{9}{\int }_{0}^{3}\frac{1}{2}\left(\left({x}^{2}{\right)}^{2}-\left(3x{\right)}^{2}\right)dx$
${M}_{y}=x-\text{bar}=\frac{2}{9}{\int }_{0}^{3}x\left({x}^{2}-3x\right)dx$

tiepidolu

The correct integrals should be
$A={\int }_{0}^{3}\left(3x-{x}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}dx=\frac{9}{2}$
${M}_{x}=\frac{2}{9}{\int }_{0}^{3}x\left(3x-{x}^{2}\right)\phantom{\rule{thinmathspace}{0ex}}dx$
${M}_{y}=\frac{2}{9}{\int }_{0}^{3}\frac{\left(3x{\right)}^{2}-\left({x}^{2}{\right)}^{2}}{2}\phantom{\rule{thinmathspace}{0ex}}dx$
since $3x>{x}^{2}$ over (0,3).

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