# Expressing infinitesimal physical quantities

In physics class, my teacher demonstrated that in polar coordinates, an infinitesimal area involving radial length dr and infinitesimal angle dθ is equal to rdr dθ, since the area is roughly a square with height dr and length rdθ.

What bothers me is that the length could have been taken as (r + dr)dθ instead, by considering the side of the square further from the origin. Clearly this would yield a different answer. I could even have considered the area of an isosceles trapezium instead.

Am I wrong? What is going on?

• $\mathrm dr(r+\mathrm dr)\mathrm d\vartheta=r\mathrm dr\mathrm d\vartheta+\mathrm dr^2\mathrm d\vartheta$, the last term is cubic in infinitesimals and thus would be negligible compared to the quadratic in infinitesimals that is in the other term. Jul 10, 2023 at 6:20

The area of a sector which subtends an angle $$\theta$$ at the centre of a circle of radius $$r$$ is $$\frac 12 r^2 \theta$$.

In your example the area you are considering is $$\frac12 (r+\delta r)^2\delta \theta - \frac12 r^2\delta \theta = (r\delta r+ (\delta r)^2/2)\delta \theta$$.
Thus the second term in the bracket becomes vanishingly smaller compared with the first term.

However, please note that all I have done is differentiated the area $$A = \frac 12 r^2 \delta \theta$$ with respect to $$r$$ from first principles to find that $$\frac {dA}{dr} = r\delta \theta$$ and then made the approximation $$\delta A \approx r\delta \theta \,\delta r$$.

I think it may be helpful to move away from the 2-D case here and just consider the simpler problem of approximating a 1-D integral from a left or right Riemann sum or a trapezoid approximation. In each case where we have discrete finite values of dx in these sums, we are getting an approximate value for the area under the curve that gets more accurate in the limit as dx approaches 0. And regardless of whether we choose left side rectangles, trapezoids, right side rectangles, in that limit as dx approaches zero the sums converge to the same answer which is the area under the curve.

In a similar sense, we could imagine chopping up the 2-D plane into different shapes and choosing some suitable function values in that small finite area and sum up the volume of all of the resulting 3-D columns, and different choices for how to divvy up the plane will have different levels of accuracy as the areas, but in the limit as the differentials approach zero they will all converge to the appropriate volume.

If we were to calculate the full area encompassed by a finite patch of the plane spanned by $$dr, d\theta$$, we would see that the region is two arcs of a circle and could calculate an area of $$rdrd\theta+dr^2d\theta/2$$. The somewhat hand-wavy answer is that the second terms is cubic in infinitesimals and therefore contributes negligibly in the limit as the infinitesimals approach zero, we only keep the term with the highest order.

The more careful and rigorous way of determining what additional factor we may need to include in an integral when changing the coordinate system is to analyze how the coordinate transformation affects areas/volumes as a function of the coordinates. The way to do that is with the Jacobian determinant as laid out in this wikipedia page: https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant