How do I solve for the value of areal density given the mass and radius?

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To solve for the value of areal density D given a solid disk with mass M=15.6kg and radius 0.25m, the approach involves integrating the surface density over the disk's area. The mass is expressed as M = ∫∫(σ dA), where σ is the areal density and dA is the area element in polar coordinates, dA = r dr dθ. The integration limits for r are from 0 to 0.25m and for θ from 0 to 2π. By setting up the integral correctly and solving for D, the same principles used for linear density can be applied. This method allows for the calculation of the areal density based on the given parameters.
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I've ran aground on my physics 4a homework.

Consider a solid disk of mass M=15.6kg and areal density =Dx3. Determine the value D if the radius of the disk is 0.25m.

I solved the problem before it: (a)Consider a two meterstick of mass M=8.2kg and linear density =Cx5. Determine the value C.

by taking the intergral of the linear density and setting it equal to the mass and plugging in 2 meters for the x and solving for C.

Changing the object to a disk from a meter stick somehow destroys the problem from me and I can't figure out what to do to get it to work. I've tried finding area pi(r)^2 using the radius that gave, and I've tried circumferance 2pi(r).

What I'm I missing?
 
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Could it be that x in Dx3 stands for the radius of the disk? In this case, you can proceed exactly as with the stick; integrating the surface density from 0 to 0.25m (and 0 to 2pi) and then solve for D.
 
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I start again...

Use the fact that the mass is the surface integral of the areal density over the whole surface. I.e.

M = \iint_{surface} \sigma dA

And use the fact that the areal density is given to you as a function of the radius to integrate in polar coordinate. Recall that in polar coordinate, the area element is dA = rdrd\theta.
 
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