Disclaimer: This was given as a homework from college but the teacher didn't teach us anything about density or mass or anything related.
A lamina has the form of the region limited by the parabola $ y = x^2 $ and the straight line $ y = x $. The density varies as the distance from the $ X $ axis.
Find the mass and center of mass.
what i could find however is that the formula of mass is the following
$$M = \int\int_R \rho(x,y)dA $$
so i tried doing something like this
$$ \int_0^1\int_y^{\sqrt(y)} ? dxdy $$
the thing is that they say the density varies as the distance from the x axis, so i don't know what to replace for the density.. is it $ x + y $?
Best Answer
The mass density varies as the distance from the x-axis implies that $\rho =Ky$ where $K$ is a constant.
Now, the total mass is given by
$$M=\int_0^1 \int_y^{\sqrt{y}} (Ky) dx dy$$
The moment about x is
$$\frac{\int_0^1 \int_y^{\sqrt{y}} x(Ky) dx dy}{M}$$
$$\frac{\int_0^1 \int_y^{\sqrt{y}} y(Ky) dx dy}{M}$$
Can you complete? Notice that the moments are independent of $K$.