# Thread: Center of Mass via Calc 3 question

1. ## Center of Mass via Calc 3 question

The question states . Find the center of mass of the lamina x^2 + y^2 less than equal to 1 if the density at any point is proportional to the square of it's distance from the origin.

My question is, what in the crap is the question stating in layman terms? I'm terrible at word problems and need to have this broken down completely.

2. The definition of the center of mass is $c_x = \frac{\int\int\int x\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$
$c_y = \frac{\int\int\int y\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$

$c_z = \frac{\int\int\int z\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$

In your case $f(x,y,z) = 1$ and $\rho(x,y,z) = x^2+y^2+z^2$

Of course, because of symmetry $c_x=c_y=0$
EG : $c_z = \frac{\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{\fra c{1}{\sqrt{2}}}\int_{x^2+y^2}^{1} z(x^2+y^2+z^2) dxdydz}{\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{\fra c{1}{\sqrt{2}}}\int_{x^2+y^2}^{1} (x^2+y^2+z^2) dxdydz} \simeq 0.749$

3. Originally Posted by JonathanEyoon
The question states . Find the center of mass of the lamina x^2 + y^2 less than equal to 1 if the density at any point is proportional to the square of it's distance from the origin.

My question is, what in the crap is the question stating in layman terms? I'm terrible at word problems and need to have this broken down completely.
$x^2+y^2 \le 1$

is the unit disc (centre at the origin), density proportional to the square of distance from origin, means that the desity is a radial function, and as the origin is the centre of the disk without integration we know that the centre of mass is the origin (by symmetry).

CB

4. Originally Posted by vincisonfire
The definition of the center of mass is $c_x = \frac{\int\int\int x\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$
$c_y = \frac{\int\int\int y\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$

$c_z = \frac{\int\int\int z\rho(x,y,z) f(x,y,z) dxdydz}{\int\int\int \rho(x,y,z) f(x,y,z)dxdydz}$

In your case $f(x,y,z) = 1$ and $\rho(x,y,z) = x^2+y^2+z^2$

Of course, because of symmetry $c_x=c_y=0$
EG : $c_z = \frac{\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{\fra c{1}{\sqrt{2}}}\int_{x^2+y^2}^{1} z(x^2+y^2+z^2) dxdydz}{\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\int_{x^2}^{\fra c{1}{\sqrt{2}}}\int_{x^2+y^2}^{1} (x^2+y^2+z^2) dxdydz} \simeq 0.749$
Where did z come from? The problem is confined to the x-y plane (the figure is a lamina).

CB

5. I mess the "lamina" part. I thought we had to find the center of mass of the region bounded by x^2+y^2 under z=1. I don't see the interest of the question if we work in the plane.