Originally Posted by

**Showcase_22** What i've tried:

$\displaystyle M= \int_0^3 \int_0^{3-x} dy \ dx = \int_0^3 3-x \ dx = \left[ 3x-\frac{x^2}{2} \right]_0^3=9-\frac{9}{2}=\frac{9}{2}$

$\displaystyle M_x=\int_0^3 \int_0^{3-x} y \ dy \ dx=\frac{1}{2} \int_0^3 (3-x)^2 \ dx=\frac{1}{2} \int_0^3 9-6x+x^2 \ dx$$\displaystyle =\frac{1}{2} \left[9x-3x^2+\frac{x^3}{3} \right]_0^3=\frac{1}{2} \left(27-27+\frac{27}{3} \right)=\frac{27}{6}=\frac{9}{2}$

$\displaystyle M_y=\int_0^3\int_0^{3-x} x \ dy \ dx=\int_0^3 3x-x^2 \ dx= \left[\frac{3x^2}{2}-\frac{x^3}{3} \right]_0^3=\frac{27}{2}-\frac{27}{3}=\frac{27}{6}=\frac{9}{2}$

$\displaystyle \overline{x}=\frac{9}{2} \left( \frac{2}{9} \right)=1$

$\displaystyle \overline{y}=\frac{9}{2} \left( \frac{2}{9} \right)=1$

This means that the centre of mass is $\displaystyle (1,1)$.

However, intuitively this doesn't seem right to me. I would expect it to be at $\displaystyle \left( \frac{3}{2}, \frac{3}{2} \right)$ since that would be the middle of the two sides (and the expression for the density is constant).

So can anyone see if $\displaystyle (1,1)$ is correct. If it isn't correct, where have I gone wrong in my integrals?

Thanks!