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 $g(x)=3x$ and g(x)=3x. Sketch R, which appropriate labels. Then set up all the integrals required do not evaluate.

My solution:

Definitions:

$A={\displaystyle {\int}_{a}^{b}(f(x)-g(x))dx}$

${M}_{x}=y-\text{bar}=\frac{1}{A}{\displaystyle {\int}_{a}^{b}\frac{1}{2}(f(x{)}^{2}-g(x{)}^{2})dx}$

${M}_{y}=x-\text{bar}=\frac{1}{A}{\displaystyle {\int}_{a}^{b}x(f(x)-g(x))dx}$

….

$A={\displaystyle {\int}_{0}^{3}({x}^{2}-3x)dx=\frac{9}{2}}$

${M}_{x}=y-\text{bar}=\frac{2}{9}{\displaystyle {\int}_{0}^{3}\frac{1}{2}(({x}^{2}{)}^{2}-(3x{)}^{2})dx}$

${M}_{y}=x-\text{bar}=\frac{2}{9}{\displaystyle {\int}_{0}^{3}x({x}^{2}-3x)dx}$

Consider the bounded region R between the graphs of $g(x)=3x$ and g(x)=3x. Sketch R, which appropriate labels. Then set up all the integrals required do not evaluate.

My solution:

Definitions:

$A={\displaystyle {\int}_{a}^{b}(f(x)-g(x))dx}$

${M}_{x}=y-\text{bar}=\frac{1}{A}{\displaystyle {\int}_{a}^{b}\frac{1}{2}(f(x{)}^{2}-g(x{)}^{2})dx}$

${M}_{y}=x-\text{bar}=\frac{1}{A}{\displaystyle {\int}_{a}^{b}x(f(x)-g(x))dx}$

….

$A={\displaystyle {\int}_{0}^{3}({x}^{2}-3x)dx=\frac{9}{2}}$

${M}_{x}=y-\text{bar}=\frac{2}{9}{\displaystyle {\int}_{0}^{3}\frac{1}{2}(({x}^{2}{)}^{2}-(3x{)}^{2})dx}$

${M}_{y}=x-\text{bar}=\frac{2}{9}{\displaystyle {\int}_{0}^{3}x({x}^{2}-3x)dx}$

tiepidolu

Beginner2022-09-08Added 8 answers

The correct integrals should be

$A={\int}_{0}^{3}(3x-{x}^{2})\phantom{\rule{thinmathspace}{0ex}}dx=\frac{9}{2}$

${M}_{x}=\frac{2}{9}{\int}_{0}^{3}x(3x-{x}^{2})\phantom{\rule{thinmathspace}{0ex}}dx$

${M}_{y}=\frac{2}{9}{\int}_{0}^{3}\frac{(3x{)}^{2}-({x}^{2}{)}^{2}}{2}\phantom{\rule{thinmathspace}{0ex}}dx$

since $3x>{x}^{2}$ over (0,3).

$A={\int}_{0}^{3}(3x-{x}^{2})\phantom{\rule{thinmathspace}{0ex}}dx=\frac{9}{2}$

${M}_{x}=\frac{2}{9}{\int}_{0}^{3}x(3x-{x}^{2})\phantom{\rule{thinmathspace}{0ex}}dx$

${M}_{y}=\frac{2}{9}{\int}_{0}^{3}\frac{(3x{)}^{2}-({x}^{2}{)}^{2}}{2}\phantom{\rule{thinmathspace}{0ex}}dx$

since $3x>{x}^{2}$ over (0,3).

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