# Lamina

• Jul 30th 2006, 07:30 PM
JaysFan31
Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Mike
• Jul 30th 2006, 07:50 PM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31
Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Mike

Below is the diamgram of the region.
In order to solve this problem you need to find the countinous density function. Which you say is proportional to the distance. I take this to mean that you are saying directly proportional right? Then you have,
$\frac{\mbox{density}}{\mbox{distance}}=\kappa$ for some constant of proportionality.
In mathematical terms,
$\frac{\delta(x,y)}{\sqrt{x^2+y^2}}=\kappa$
Thus,
$\delta(x,y)=\kappa \sqrt{x^2+y^2}$

In order to find the center of mass you first need to find the mass. Which in this case is,
$\int_R \int \delta(x,y) dA$
Thus,
$\int_R \int \kappa \sqrt{x^2+y^2}dA$
I believe it is easier by converting to polar cordinates.
Thus,
$\int_0^{2\pi} \int_6^7 \kappa r^2 dr \, d\theta$
Thus,
$\int_0^{2\pi} \frac{1}{3}\kappa r^3 |_6^7= 84.\bar 6\pi \kappa$--->Mass
----------
I will continue later.
• Jul 30th 2006, 09:52 PM
CaptainBlack
Quote:

Originally Posted by JaysFan31
Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Mike

I will rewrite the definition of the region more clearly for you:

$
36 \le x^2+y^2 \le 49, \ \ y \ge 0
$
,

which is the top half of the shaded region in hacker's diagram.

RonL

of mass would of course have been at $(0,0)$ by symmetry)
• Jul 31st 2006, 05:35 AM
CaptainBlack
Quote:

Originally Posted by JaysFan31
Any help on this problem would be greatly appreciated. It deals with multiple integration:

Find the centre of mass (x,y) of the lamina which occupies the region:
36 less than or equal to (x^2) + (y^2) less than or equal to 49
y greater or equal to 0
if the density at any point is proportional to the distance from the origin.

Mike

Let the density $\rho(r)=\kappa r$, theb=n the mass of the
lamina is:

$
M=\int_{\theta=0}^{\pi} \int_{r=6}^7 \rho(r)\ r\ dr d\theta$
$=
\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^2dr\ d\theta
$

Now the centre of mass is:

$
\bold{R}=\frac{1}{M} \int_A \rho(r)\ \bold{r} dA
$

where $\bold{r}=r(\cos(\theta),\sin(\theta))$.

So:

$
\bold{R}=\frac{1}{M}\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^3\ (\cos(\theta),\sin(\theta)) dr\ d\theta
$
.

You should be able to complete these integrals yourself.

RonL
• Jul 31st 2006, 02:26 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank

Now the centre of mass is:

$
\bold{R}=\frac{1}{M} \int_A \rho(r)\ \bold{r} dA
$

Is that a line integral? You can calculate center of mass with line integral?
• Jul 31st 2006, 10:25 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Is that a line integral? You can calculate center of mass with line integral?

It's an area integral.
• Aug 1st 2006, 09:56 AM
JaysFan31
OK. I was able to get the centre of mass for x (it equals zero). How does the centre of mass for y differ? It does not equal zero. What changes in the integration process?

Thanks.

Mike
• Aug 1st 2006, 10:55 AM
CaptainBlack
Quote:

Originally Posted by JaysFan31
OK. I was able to get the centre of mass for x (it equals zero). How does the centre of mass for y differ? It does not equal zero. What changes in the integration process?

Thanks.

Mike

What changes is that $\sin(\theta)$ is positive over the entire
range of integration (that is $(0,\pi)$), so the integral over $\theta$
does not vanish as it does for the $\cos(\theta)$ in the integral
for the $x$ coordinate.

RonL
• Aug 1st 2006, 08:43 PM
JaysFan31
I can't figure out the integral for the y-coordinate. Quick assistance would be greatly appreciated.

Is it the integration of (r^3)sin(theta)sin(theta) drdtheta with the same bounds?
• Aug 1st 2006, 09:13 PM
CaptainBlack
Quote:

Originally Posted by JaysFan31
I can't figure out the integral for the y-coordinate. Quick assistance would be greatly appreciated.

Is it the integration of (r^3)sin(theta)sin(theta) drdtheta with the same bounds?

You want:

$
\bold{R_y}=\frac{1}{M}\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^3\ \sin(\theta)\ dr\ d\theta
$

RonL