1. ## Finding a centroid

Find the centroid of the rectangular region cut from from the first quadrant by the line $x+y=3$.
What i've tried:

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

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

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

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

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

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

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

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

Thanks!

2. Originally Posted by Showcase_22
What i've tried:

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

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

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

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

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

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

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

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

Thanks!
It's only in the centre of the object if the object is symmetrical about that axis. You can see that the triangle is symmetrical if you look at it from the point of view of it's hypotenuse. And the coordinate (1,1) is indeed in its centre with respect to that axis. So all is well. You wouldn't expect it to be in the centre on any other axis because it isn't symmetrical on any other axis. (1,1) is fine.