# Find Mass and C.O.M using a line integral

1. May 14, 2013

### Smazmbazm

1. The problem statement, all variables and given/known data

Using line integrals, find the mass and the position of the center of mass of a thin wire in the shape of a half-circle $x^{2} + y^{2} = r^{2}, x ≥ 0$ and $-r ≤ y≤ r$ if the linear density is $ρ(x,y) = x^{2} + y^{2}$

The mass is given by the integral of the density along the curve while the center of mass is defined as

$\frac{∫_{C} x ρ(x,y)ds}{∫_{C}ρ(x,y)ds}$

for the x co-ordinate and similarly for the y co-ordinate.

3. The attempt at a solution

My attempt was to substitute $x^{2} = y^{2} - r^{2}$ and $y^{2} = x^{2} - r^{2}$ into the density function to get $ρ(x,y) = x^{2} + y^{2} - 2r^{2}$ then do the double integral

$∫^{x}_{0}∫^{r}_{-r} x^{2} + y^{2} - 2r^{2} dy dx$

from that I got an answer of $\frac{2rx^{2}}{3}$

EDIT: I get $\frac{2rx^{2}}{3} - \frac{10r^3x}{3}$ rather then $\frac{2rx^{2}}{3}$

Last edited: May 14, 2013
2. May 14, 2013

### LCKurtz

Stop right there! Those are line integrals and you don't evaluate them with double integrals. Do the line integrals. Polar coordinates might be a good idea.

3. May 14, 2013

### ParoxysmX

Hah, physics 211 I assume?

4. May 15, 2013

### Smazmbazm

Physics 211 indeed, ParoxysmX =]. LCKrutz, ok so,

$x = r cos(t), y = r sin(t)$

$ds = \sqrt{((dx/dt)^{2} + (dy/dt)^{2})} = \sqrt{(r^{2}sin^{2}(t) +r^{2}cos^{2}(t))}dt = \sqrt{r^2}dt = r dt$

Then substitute x and y values into the density function to get

$Mass = ∫_{C}(r^{2}cos^{2}(t) + r^{2}sin^{2}(t))(r)dt$
$Mass =∫_{C}r^{3}(cos^{2}(t)+sin^{2}(t))dt = ∫_{C}r^{3}dt$
$Mass =r^{3}t |^{\pi}_{0} = r^{3}\pi$

Correct?

Then for the x and y, do the above process but multiply by $r cost$ and $r sint$ and divide by $r^{3}\pi$ to get x and y which are $x = 0$ and $y = 2r^{4}/r^{3}\pi$.