Mass of 2 dimensional object - line integral

• Liferider
I still can't figure it out though...\bar{x}=\frac{1}{2r^3}\int^{0.5pi}_{0}r3(cos3(t)+cos2(t)sin(t))dt[/QUOTE]In summary, we are trying to find the mass and coordinates for the center of mass of a quarter circle thin wire. We can use the circle equation and mass density equation to help us solve the problem. After some calculations, we find that the mass is equal to 2r^2. To find the x coordinate of the center of mass, we use the formula \bar{x} = \frac{1}{M}\int x\,dm, and after some adjustments, we find

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

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.

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

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.

Liferider said:
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...
As I said before, $ds= \sqrt{(x')^2+ (y')^2}dt= rdt$
The density is given by x+ y= r(cos(t)+ sin(t)) so the mass is given by the intergral
$$r^2\int_{t=0}^{\pi/2} (cos(t)+ sin)(t))dt= r^2(sin(t)- cos(t))_0^{\pi/2}= r^2((1- 0)- (0- 1))= 2r^2$$ 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 $r^2$.

Thanks for the replies :-)

What is the concept of "mass of 2 dimensional object - line integral"?

The concept of "mass of 2 dimensional object - line integral" is a mathematical approach used to calculate the mass of a two-dimensional object by taking the line integral of its density function over its domain.

How is the line integral used to calculate the mass of a 2 dimensional object?

The line integral is used to calculate the mass of a 2 dimensional object by integrating the density function of the object along a chosen path or curve. This involves breaking down the object into infinitesimal line segments and summing up their masses to find the total mass of the object.

What factors affect the accuracy of calculating the mass using line integral?

The accuracy of calculating the mass using line integral depends on the choice of path or curve along which the integration is performed, as well as the accuracy of the density function used. It is also important to ensure that the object is broken down into small enough segments to minimize error.

What are some real-life applications of calculating mass using line integral?

Calculating mass using line integral has various applications in physics and engineering, such as calculating the mass of a thin wire, a thin plate, or a fluid flow. It is also used in computer graphics and animation to calculate the mass of 2D objects for realistic simulations.

What is the difference between calculating the mass of a 2 dimensional object using line integral and using traditional methods?

The traditional method of calculating the mass of a 2 dimensional object involves finding the area under the curve of its density function, while the line integral method involves integrating the density function along a chosen path. The line integral method allows for more flexibility in the choice of path and can be used for irregularly shaped objects.