1. ## Center of mass

A lattice is on the portion of the disk $x^2+y^2 \leq 1$ in the first quadrant. Find its center of mass if the density at any point is proportional to its distance with the $x$ axis.
My attempt : I first calculated the total mass of the lattice which gave me $\frac{k}{3}$, where $k$ is the constant of proportionality.
Then I found out that the center of mass is located at $\left( \frac{3}{8},\frac{3}{16} \right)$.

I just want to know if this is the correct result because I find it quite strange.

2. I agree with your total mass and the x coordinate of the c of m. How did you arrive at the y coord.?

3. Originally Posted by Danny
I agree with your total mass and the x coordinate of the c of m. How did you arrive at the y coord.?
I've made many errors and now I'm not able to read my notes. Is $y=\frac{3\pi}{16}$?
$y=\frac{k \int_0^{\pi/2} \int_0^{1} \rho ^3 \cos ^2 (\theta) d\rho d\theta}{\frac{k}{3}}=\frac{3\pi}{16}$.

4. Originally Posted by arbolis
I've made many errors and now I'm not able to read my notes. Is $y=\frac{3\pi}{16}$?
$y=\frac{k \int_0^{\pi/2} \int_0^{1} \rho ^3 \cos ^2 (\theta) d\rho d\theta}{\frac{k}{3}}=\frac{3\pi}{16}$.
Yes, that's what I obtained. Side note - most people use $r$ and not $\rho$ for the radius in 2D problems.

5. Originally Posted by Danny
Yes, that's what I obtained. Side note - most people use $r$ and not $\rho$ for the radius in 2D problems.
You're right. I'm the odd one.

6. Two things I notice immediately are that the figure and density function are both symmetric in x and y. The x and y coordinates of the center of mass must be the same.

7. Originally Posted by arbolis
A lattice is on the portion of the disk $x^2+y^2 \leq 1$ in the first quadrant. Find its center of mass if the density at any point is proportional to its distance with the $\color{red}{x}$ axis.
My attempt : I first calculated the total mass of the lattice which gave me $\frac{k}{3}$, where $k$ is the constant of proportionality.
Then I found out that the center of mass is located at $\left( \frac{3}{8},\frac{3}{16} \right)$.

I just want to know if this is the correct result because I find it quite strange.
Originally Posted by HallsofIvy
Two things I notice immediately are that the figure and density function are both symmetric in x and y. The x and y coordinates of the center of mass must be the same.
I beg to differ.