Computing center of mass

Win_odd Dhamnekar

Win_odd Dhamnekar

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Find the center of mass of the region R with the given density function δ(x,y)\displaystyle \delta(x,y)


1)R = {(x,y) : y ≥ 0, x ≥ 0 , 1 ≤ x2 + y2 ≤ 4 }, δ(x,y)=x2+y2\displaystyle \delta(x,y) = \sqrt{x^2 + y^2}

How to answer this question?

My attempt to answer this question:

M=∫02∫0x(x2+y2)32dydx=10.0348925582M = \displaystyle\int_0^2 \displaystyle\int_0^x (x^2+ y^2)^{\frac32}dy dx = 10.0348925582
Is this computation of M correct? How to compute M in cylindrical coordinates?

How to compute center of mass in any coordinates?(spherical, cylindrical, cartesian)
 
A hint: the center of mass has two coordinates, so computing one integral isn't going to be enough.
 
Dhamnekar Winod said:
Find the center of mass of the region R with the given density function δ(x,y)\displaystyle \delta(x,y)


1)R = {(x,y) : y ≥ 0, x ≥ 0 , 1 ≤ x2 + y2 ≤ 4 }, δ(x,y)=x2+y2\displaystyle \delta(x,y) = \sqrt{x^2 + y^2}

How to answer this question?

My attempt to answer this question:

M=∫02∫0x(x2+y2)32dydx=10.0348925582M = \displaystyle\int_0^2 \displaystyle\int_0^x (x^2+ y^2)^{\frac32}dy dx = 10.0348925582
Is this computation of M correct? How to compute M in cylindrical coordinates?

How to compute center of mass in any coordinates?(spherical, cylindrical, cartesian)
Click to expand...
Assuming you did the integration correctly (I didn't check) you have found the total mass, not the center of mass. To do that you need
xcm=∫Axδ dA∫δ dAx_{cm} = \dfrac{ \int_A x \delta ~ dA }{ \int \delta ~ dA }

and similarly for y.

To do this in polar coordinates you want ∫12∫0π/2...r dθ dr\int_1^2 \int_0^{\pi/2} ... r ~ d \theta ~ dr for this example.

-Dan

Addendum: You have the wrong integration limits. x and y start at 1, not 0. Polar coordinates are definitely the way to go.

It would be something like
∫12∫1−x24−x2... dy dx\int_1^2 \int_{\sqrt{1 - x^2}}^{\sqrt{4 - x^2}} ... ~ dy ~ dx
 
Last edited:
The center of mass of the region R is (x‾,y‾)\displaystyle (\overline{x}, \overline{y}) where,

1654699248178.png

I computed M (the denominator) in the computaion of x‾,y‾\displaystyle \overline{x}, \overline{y} .
 
Cartesian Coordinate.

M=∫01∫1−x24−x2x2+y2  dy dx+∫12∫04−x2x2+y2  dy dx\displaystyle M = \int_{0}^{1}\int_{\sqrt{1-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2 + y^2} \ \ dy \ dx + \int_{1}^{2}\int_{0}^{\sqrt{4-x^2}} \sqrt{x^2 + y^2} \ \ dy \ dx

Polar Coordinate.

M=∫0π/2∫12r2 dr dθ\displaystyle M = \int_{0}^{\pi/2}\int_{1}^{2} r^2 \ dr \ d\theta
 
So, the center of mass of the region R is (x‾,y‾)=(4514π,4514π)\displaystyle (\overline{x}, \overline{y}) =\left(\frac{45}{14\pi}, \frac{45}{14\pi}\right)
 
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