Moment of Inertia about the line x=y

[Jadehaan]

Homework Statement

Let \Omega represent a two dimensional material region who density is given by \rho(x,y). Establish an integral formula for the moment of inertia of the material region about the line y=x.

Homework Equations

The moment of inertia about the x axis: \int\intx²\rho(x,y)dA

The Attempt at a Solution

I am attempting to transform the moment of inertia about the x-axis to the line x=y, how would I go about doing this?

Thanks for any help, Jim.

 

[Dick]

The moment of inertia is the integral of r^2 rho(x,y) dA where r is the distance from the point (x,y) to the axis. That means the expression you gave is the moment of inertia about the y axis, not the x axis. Now you have to replace r in the integral with the distance from a point (x,y) to the line x=y. How many ways do you know to find the distance from a point to a line?

 

[Jadehaan]

Thanks for your help. You were right, I mistyped the formula for the moment about the x axis. Using points (x,y) and (y,x) I calculated r²=2x²-4xy+2y² So the formula would be the integral of this multiplied by rho dA.

 

[Dick]

That seems close, but aren't you off by a factor of four there? The distance from (1,0) to the line x=y is sqrt(2)/2. Isn't it? So that makes r^2 for (1,0) r^2=1/2.

 

[Jadehaan]

I see. I was using the wrong distance formula. So using the formula for the distance between a point (m,n) and a line Ax+By+c=0 the distance is r= |Am+Bn+C|/ sqrt(A²+B²)
Using the point (x,y) and the line x-y=0 I calculated the distance r=|x-y|/sqrt(2)
Does this sound correct?

 

[Dick]

Yes, that sounds correct. So r^2=(x-y)^2/2.

 

[Jadehaan]

Yep, thanks so much Dick.