1. ## Question about a center of mass/integrals problem

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!

2. 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

$\chi$ $\sigma$