Question about a center of mass/integrals problem

**steph3824** · July 11th 2010, 02:10 PM

Set up the integrals to find the center of mass of an equilateraly triangle with density ρ(x,y)=K√ (x²+y²), with sides of length two and the midpoint of one side is the origin.

Ok first question..I know that the limits on the x integral are from -1 to 1, but I am not sure what to put for y's limits. I am not sure if it would be 0 to 2-2x, 0 to 2+2x, or 2+2x to 2-2x, so I need some help with this please.

Also, I know that I need to set up the integral for mass first. But after this, in my book, sometimes integrals are set up for x bar and y bar, and then in other cases integrals are set up for M_y and M_x. So my question is, which two should I do in the case of a triangle? Does it even make a difference?

Thanks so much for your help!

**chisigma** · July 12th 2010, 09:15 AM

The triangle whose coordinates of the center of mass You have to compute is illustrated in the figure...

The computation is performed in two steps...

a) first computing the mass m of the triangle...

$\displaystyle m= \int \int _{A} \rho (x,y)\ dx\ dy$ (1)

b) then computing the coordinates of the center of mass...

$\displaystyle x_{m} = \frac{1}{m}\ \int \int _{A} x\ \rho (x,y)\ dx\ dy$

$\displaystyle y_{m} = \frac{1}{m}\ \int \int _{A} y\ \rho (x,y)\ dx\ dy$ (2)

For the function $\rho (x,y)$ we can set k=1 so that is $\rho = \sqrt {x^{2} + y^{2}}$ and the integral (1) becomes...

$\displaystyle m= \int_{0}^{\sqrt{3}} \int_{-1 + \frac{x}{\sqrt{3}}}^{1 - \frac{x}{\sqrt{3}}} \sqrt{x^{2} + y^{2}}\ dx\ dy$ (3)

... and integrals (2)...

$\displaystyle x_{m}= \frac{1}{m}\ \int_{0}^{\sqrt{3}} \int_{-1 + \frac{x}{\sqrt{3}}}^{1 - \frac{x}{\sqrt{3}}} x\ \sqrt{x^{2} + y^{2}}\ dx\ dy$

$\displaystyle y_{m}= \frac{1}{m}\ \int_{0}^{\sqrt{3}} \int_{-1 + \frac{x}{\sqrt{3}}}^{1 - \frac{x}{\sqrt{3}}} y\ \sqrt{x^{2} + y^{2}}\ dx\ dy$ (4)

Kind regards

$\chi$ $\sigma$

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