Wikipedia has a helpful elementary discussion of resistance [unit = Ohm] of an object as it relates to the resistivity [unit = Ohm-cm] of a material. The easy relationship is that the resistance in Ohms of a 1-cm cube of material is equal numerically to its resistivity in Ohm-cm. Resistance = resistivity x length / area in compatible units.

It thus makes sense to speak of a **material** having a resistivity that is [or, as in the case of the OP, is not] constant. Thus, in the formula R(p) = 2*sqrt(p), R refers to the resistivity of the material. After you've specified the material's shape, i.e. created an **object**, it's meaningful to speak of resistance.

Again, the distinction is that resistivity is a property of a material, independent of its shape; resistance is a property of an object. Resistance depends on the object material's resistivity and on the object's shape.

I was attempting to recreate a problem from an EE final I had 1.5 years ago based on memory, but I realized after I submitted that I had some minor problems in the statement. One which causes the actual calculation to be slightly more difficult than I anticipated (harder integral), and now.... units.

I would argue that the differential volume elements have shape and are objects, but this would not be helpful, because integration involves more than just summing stuff together - it also involves a change in units due to multiplication of differential elements of whatever units belong to the variable of integration. Thus if you intend to use something as part of an integrand, such that after integration it becomes what you originally had, what's in the integrand is actually a density of the original quantity vs. being portions of the original quantity.

In essence, bonanova is correct.

**Edited by mmiguel, 06 September 2012 - 07:48 AM.**