Integration Application: Linear Density Question

[Bwave]

I recently was marked down on my homework when solving a problem relating to linear density. I understand the process of computing the mass, but in my mind it doesn't make sense. I would appreciate if someone could explain it to me.

 

The problem similar to as follows:

Calculate the mass of a 3 meter rod whose linear density is calculated using the formula p(x) = sin(3x)+5 kilograms per meter.

 

I understand you setup an integral w.r.t. x but in my mind the problem is missing a component. Wouldn't the mass of the rod vary depending on the radius of the rod? I understand that the density is linear, but if the rod was 30 meters in radius, wouldn't it be greater than if the rod had a .1 meter radius?

 

My final answer ended up including a variable R because I thought the mass would change depending on the thickness. I simply found the cross-sectional area of the rod which was just a circle: piR^2. I then multiplied the circle times the density equation. I then computed the volume by adding up all the circles from the beginning of the rod to the end of the rod. It ended up being 2piR^2 times the integral from 0 to 3 of (x(sin(3x)+5))dx. The correct answer was simply a value and I am a little bit confused by this.

 

Could anyone provide a reasonable explanation as to why the radius is not taken into account in linear density problems?

[Schrödinger's hat]

Could anyone provide a reasonable explanation as to why the radius is not taken into account in linear density problems?

 

Look at the units.

It's not mass per unit volume, it's mass per unit length. The radius is already accounted for.

[Bwave]

Thank you very much. It's the small things that usually are of utmost importance.