Mass of 2 dimensional object - line integral

Homework Statement

Find the mass and the coordinates for the center of mass of a thin wire formed like a quarter circle.

Homework Equations

Circle equation: x2+y2=r2
Mass density: rho=x+y

The Attempt at a Solution

I know that: x2+y2=cos2(t)+sin2(t)=1
x=r*cos(t)
y=r*sin(t)

The quarter circle is described by the vector equation:
r(t)=r*cos(t)i+r*sin(t)j; 0<t<1/2*pi
The density function becomes:
rho=x+y=r*cos(t)+r*sin(t)=r(cos(t)+sin(t))

The line integral that I try to use: (C is the quarter circle wire)
M=∫$_{C}$(rho(x,y)*ds)=∫$^{1/2*pi}_{0}$(rho(x,y)*r'(t)*dt)
=r2∫$^{1/2*pi}_{0}$(cos2(t)-sin2(t))*dt

This ends up in M=0.... FFFUUUU! The right answer is supposed to be 2r2

HallsofIvy
Homework Helper
You appear to be arguing that $r'= r(-sin(t)i+ cos(t)j)dt$ and then taking the dot product with r cos(t)i+ r sin(t)j. But x+ y is NOT a vector. Instead you need to use the scalar differential, $ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt$.

Ahh, noob error Thanks :-) I'll try that

Ok, I found the correct solution for the mass, but now I keep hitting the wall with the center of mass...

The way I have tried to find it, is by two integrals, one for the x coordinate and one for the y coordinate.

$\bar{x}$=$\frac{1}{M}$$\int$x(t)*$\rho$(x(t),y(t))*dt

M=2r2
x=r*cos(t)
y=r*sin(t)
$\rho$=x+y

I regarded r as constant, so its a single integral. Should this be a double integral instead, with r as a variable? The answer I found with this integral was:

$\bar{x}$=$\frac{1}{8}$($\pi$+1)

It should be $\bar{x}$=$\frac{r}{8}$($\pi$+2)

I fairly new to vector calculus, double and triple integrals +, so please bear with me.

HallsofIvy
Homework Helper
Well, when I do the calculation, I get $(r/8)(\pi+2)$. Unless you show exactly what you did, we can't say exactly what you did wrong.

Hmm

$\bar{x}$=$\frac{1}{2r^2]}$$\int$$^{0.5pi}_{0}$r2(cos2(t)+cos(t)sin(t)*dt

When I factor out r2, it gets canceled by the mass. My error seems to be here someplace...

I wonder why I can't make a double integral with x and y as variables. Just insert the limits of t in the x and y equations to get the limits for x and y, like this:

x0=r*cos(0)=r
x1=r*cos(0.5pi)=0

y0=r*sin(0)=0
y1=r*sin(0.5pi)=r

Then the double integral becomes:

$\bar{x}$=$\frac{1}{M}$$\int$$^{0}_{r}$$\int$$^{r}_{0}$(x*(x+y))dydx=$\frac{1}{2r^2}$$\int$$^{0}_{r}$$\int$$^{r}_{0}$(x2+xy)dydx

I find: -$\frac{7r^2}{24}$

vela
Staff Emeritus
Homework Helper
The latter method doesn't work because x and y aren't independent. You're integrating over the rectangle [0,r]x[0,r] whereas you only want to integrate along the wire.

When you calculated the mass, you did some integral which amounts to
$$M = \int dm$$where dm corresponds to whatever expression you had in your integral. For the center of mass, you do a similar integral with a factor of x or y thrown in
$$\bar{x} = \frac{1}{M}\int x\,dm.$$Compare what you wrote down when calculating M and your integral when calculating $\bar{x}$. The only difference should be that the latter has a factor of r cos θ in it.

HallsofIvy
$\bar{x}$=$\frac{1}{2r^2]}$$\int$$^{0.5pi}_{0}$r2(cos2(t)+cos(t)sin(t)*dt
As I said before, $ds= \sqrt{(x')^2+ (y')^2}dt= rdt$
$$r^2\int_{t=0}^{\pi/2} (cos(t)+ sin)(t))dt= r^2(sin(t)- cos(t))_0^{\pi/2}= r^2((1- 0)- (0- 1))= 2r^2$$ so you have that right.
Your error is in the "r^2" you have with the integral on the right. There is an "r" in ds= r dt, there is an "r" in x+ y= r(cos(t)+ sin(t)), and there is an "r" in x= r cos(t). You should have $r^3$ multiplying the integral, not $r^2$.