plane bounded by the lines y=x, y=4x and the hyperboals xy=1 and xy=9. find the center of mass of D.

My thought: the problem is not given any density functions, What do I have to do? First, I have to find the area of the two regions first?>

Printable View

- May 17th 2009, 01:47 AMyzc717center of mass
plane bounded by the lines y=x, y=4x and the hyperboals xy=1 and xy=9. find the center of mass of D.

My thought: the problem is not given any density functions, What do I have to do? First, I have to find the area of the two regions first?> - May 17th 2009, 04:18 AMthe_doc
If it doesn't give you any density function then I expect that you're meant to assume it's a constant.

So I'll start you off:

Let the C.M be given by $\displaystyle (\mu_x , \mu_y)$ so then

$\displaystyle \mu_x = \frac{\int \int_D x\, \mathrm{d}x \, \mathrm{d}y}{\int \int_D \, \mathrm{d}x \, \mathrm{d}y}$

and

$\displaystyle \mu_y = \frac{\int \int_D y \, \mathrm{d}x \, \mathrm{d}y}{\int \int_D \, \mathrm{d}x \, \mathrm{d}y}$ .

So all you need to do is set your limits correctly to carry out the double integral over D.

I trust you know how to do this. If your still having difficulty let me know. - May 17th 2009, 05:49 AMthe_doc
In response to your PM, here's how to deal with the limits.

Firstly it is definitely one region. Please plot the four lines to see that they intersect to create a four sided region.

Ok, so the easiest way to do this, in my opinion, is to carry out a variable transformation of the form $\displaystyle (x,y) \rightarrow (u(x,y),v(x,y)) $. Namely,

$\displaystyle u = \frac{y}{x}$ and $\displaystyle v = xy$ .

Consequently in terms of u and v we have that

$\displaystyle x^2 = \frac{v}{u}$ and $\displaystyle y^2 = uv$ .

By doing this the limits for $\displaystyle u$ will clearly be 1 to 4 and those for v will be 1 to 9. So you have that

$\displaystyle \int \int_D \, \mathrm{d}x \, \mathrm{d}y = \int_{1}^{9} \int_{1}^{4} |J| \, \mathrm{d}u \, \mathrm{d}v $

where |J| is the Jacobian for this transformation given by:

$\displaystyle \left|\begin{array}{c c}

\partial_u \, x & \partial_v \,x \\

\partial_u \, y & \partial_v \,y

\end{array}\right|$

where I've used the notation such that $\displaystyle \partial_u \, x = \frac{\partial x}{\partial u}$.

Is it OK from there? - May 17th 2009, 06:07 AMyzc717
- May 17th 2009, 06:41 AMthe_doc
Hello again!

I'm happy to help so no worries about the time. :)

So here we have the graph.

As you can see there are four lines which bound a single four sided region which is the region D.

The steepest straight line is $\displaystyle y = 4x$

Other straight line is $\displaystyle y = x$

Lower curve is $\displaystyle y = \frac{1}{x}$

Upper curve is $\displaystyle y = \frac{9}{x}$ - May 17th 2009, 06:46 AMthe_doc
It shouldn't be that complicated.

Remember you are doing partial derivatives so you should have

$\displaystyle J = \left|

\begin{array}{c c}

\frac{v}{2x}& \frac{u}{2x} \\

&\\

-\frac{v}{2yu^2} & \frac{1}{2yu}

\end{array}\right|$ .

I should also mention when finding the partial derivatives don't take the square root of $\displaystyle x^2$ just differentiate directly from that expression. For example:

$\displaystyle x^2 = uv $

so

$\displaystyle 2x \frac{\partial x}{\partial u} = v$ .

The final result for the Jacobian should be $\displaystyle \frac{1}{2u}$. - May 17th 2009, 07:20 AMthe_doc
Oops - I just realised what you mean by the two regions. If this is what your question is referring to then there isn't any work to do as the CM would be at the origin i.e. (0,0) by symmetry!

This is why I was assuming your question meant only in the first quadrant.