# Integration problem, center of mass

• February 2nd 2009, 09:27 AM
smorri28
Integration problem, center of mass
Hi could someone please help me set up the integrals for this problem. I am trying to find the center of mass. Thank you

1. for the region bounded by the parabola x=y^2-y and y=x
• February 2nd 2009, 10:29 PM
DeMath
In your case, I think you need to use these formulas to find the coordinates of the center of mass:

abscissa $\xi = \frac{{\int_0^2 {y \cdot x\left( y \right)dy} }}
{{\int_0^2 {x\left( y \right)dy} }}$
,

ordinate $\eta = \frac{{\frac{1}{2}\int_0^2 {{x^2}\left( y \right)dy} }}{{\int_0^2 {x\left( y \right)dy} }}$.

$x\left( y \right) = y - \left( {{y^2} - y} \right) = 2y - {y^2}$

$2y - {y^2} = 0 \Leftrightarrow y\left( {2 - y} \right) = 0 \Rightarrow \left[ \begin{gathered}{y_1} = 0, \hfill \\{y_2} = 2. \hfill \\ \end{gathered} \right.$

$\int_0^2 {x\left( y \right)dy} = \int_0^2 {\left( {2y - {y^2}} \right)dy} = \left. {\left( {{y^2} - \frac{{{y^3}}}{3}} \right)} \right|_0^2 = 4 - \frac{8}{3} = \frac{4}{3}.$

$\int_0^2 {y \cdot x\left( y \right)dy} = \int_0^2 {y\left( {2y - {y^2}} \right)dy} = \int_0^2 {\left( {2{y^2} - {y^3}} \right)dy} =$

$= \left. {\left( {\frac{{2{y^3}}}{3} - \frac{{{y^4}}}{4}} \right)} \right|_0^2 = \frac{{16}}{3} - \frac{{16}}{4} = \frac{{16}}{{12}} = \frac{4}{3}.$

$\int_0^2 {{x^2}\left( y \right)dy} = \int_0^2 {{{\left( {2y - {y^2}} \right)}^2}dy} = \int_0^2 {\left( {4{y^2} - 4{y^3} + {y^4}} \right)dy} =$

$= \left. {\left( {\frac{{4{y^3}}}{3} - {y^4} + \frac{{{y^5}}}{5}} \right)} \right|_0^2 = \frac{{32}}{3} - 16 + \frac{{32}}{5} = \frac{{160 - 240 + 96}}{{15}} = \frac{{16}}{{15}}.$

Finally

$\xi = \frac{4}{3}:\frac{4}{3} = 1$ and $\eta = \left( {\frac{1}{2} \cdot \frac{{16}}{{15}}} \right):\frac{4}{3} = \frac{8}{{15}}:\frac{4}{3} = \frac{2}{5}$.