Calculus lamina help

[hjohngunther]

Consider the lamina ( with uniform density .01m) having the shape of the region Q in the xy-plane: 1≤x^2+y^2≤4. The density at each point P varies directly with the distance from P to the origin, with density at the point P(2,0,0) being 10kg/m3. Find the polar moment of inertia Io

I have a big test so any help greatly appreciated, thanks

 

[HallsofIvy]

Consider the lamina ( with uniform density .01m)]

surely this is not what you meant to say! First ".01 m" is NOT a meaure of density and second, you say in the next sentence that the densithy is NOT uniform but varies. I am going to assume you meant uniform depth or thickness of 0.01 meter

having the shape of the region Q in the xy-plane: 1≤x^2+y^2≤4. The density at each point P varies directly with the distance from P to the origin, with density at the point P(2,0,0) being 10kg/m3. Find the polar moment of inertia Io

I have a big test so any help greatly appreciated, thanks.

You have a big test covering this material and you have no idea at all how to approach it? That is not good!

Do you recognize "\(\displaystyle 1\le x^2+ y^2\le 4\) as being the region between the two circles with center at the origin and radii 1 and 2? That's important because that tells you that converting to polar coordinates will make the problem easier. In polar coordinates, the "differential of area" is \(\displaystyle r drd\theta\). Since the lamina (what is "lamina" spelled backwards?) has a uniform depth of 0.01 m, the volume of such a small region is \(\displaystyle .01 r drd\theta\) and the "polar moment of inertia" is given by the density times \(\displaystyle r^2 dV= 0.01 r^3 drd\theta\) times the density. Since the density "varies directly with distance from P to the origin", the denisity is kr for some constant k. With "density at the point (2, 0, 0) being 10 kg/m^3" and (2, 0, 0) being distance 2 from the origin, we must have k(2)= 10 so that k= 5 and the density at distance r from the origin is 5r and the polar moment of inertia is \(\displaystyle 0.05 r^3drd\theta\).

Integrate that with r going from 1 to 2, \(\displaystyle \theta\) from 0 to \(\displaystyle 2\pi\).