Moment of Inertia perpendicular to lamina

[ChemistryJack]

Hi, one of my vac work questions is:

A lamina has density d(x,y) = x^2 + y^2 and is defined by -2<x<2 ; -3<y<3. Calculate the moment of inertia about an axis perpendicular to the lamina through the point (1,1).

I'm confident that I find Ix which is equal to the double integral of the density multiplied by y^2 and Iy which is equal to the double integral of the density multiplied by x^2 and add them. However, it's the limits that I'm unsure of, because for a similar question before, I integrated using the dimensions of the lamina, and my tutor said my limits should be different, I know it has something to do with the fact the moment of inertia is taking place at a given point. Could someone please help me with my limits?

Thanks a lot, also I apologise if it's difficult to read because I don't know how to write the symbols etc. properly online.

 

[gneill]

Presumably you can use the boundaries of the lamina as the integration limits provided that you make the appropriate allowance for the distance calculations. You need the square of the distance between the axis of rotation and the given mass element of the lamina. Essentially it's a coordinate transformation for the radius vectors.

For a point (x,y), the square of the distance from your axis of rotation at (1,1) is

$$r^2 = (x - 1)^2 + (y - 1)^2$$