Deriving An Integration Formula Over Polar Rectangles

We can describe a polar rectangle as in the figure below, with \(R= \{(r,θ)|a \le r \le b, \alpha \le \theta \le β \}\).

Consider a function \(f(r,θ)\) over a polar rectangle \(R\). We divide the interval \([a,b]\) into m subintervals \([r_{i-1},r_i]\) of length \(\Delta r=(b-a)/m\) and divide the interval \([α,β]\) into \(n\) subintervals \([θ_{j-1},θ_j]\) of width \(\Delta θ=(β−α)/n\).

(a) A polar rectangle \(R\). (b) divided into subrectangles \(Rij\). (c) Close-up of a subrectangle. Image from OpenStax.org.

The area \(\Delta A\) of the polar subrectangle \(R_{ij}\) is: approximately:

$$\Delta A = \frac{1}{2} \Delta r (r_{i-1} \Delta \theta + r_i \theta)$$

This is because the subrectangle looks like a trapezoid.The smaller arc length is \(r_{i-1} \Delta \theta\), the larger arc length is \(r_i \Delta \theta\), and both of these arc length are separated by a distance of \(\Delta r\). Let \(r^*_{ij} = \frac{1}{2} (r_{i-1} + r_i)\), we have \(\Delta A = r^*_{ij} \Delta r \Delta \theta\).

Going back to \(f(r, \theta)\). The function \(f\) is defined over the region \(R\). The polar volume of the thin box above \(R_{ij}\) is:

$$f(r^*_{ij}, \theta^*_{ij}) \Delta A = f(r^*_{ij}, \theta^*_{ij}) r^*_{ij} \Delta r \Delta \theta $$

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as:

$$\sum^m_{i=1} \sum^n_{j=1} f(r^*_{ij}, \theta^*_{ij}) r^*_{ij} \Delta r \Delta \theta $$

Now we can define the polar volume as the limit of double Riemann sum:

$$V = \lim_{m,n \to \infty} \sum^m_{i=1} \sum^n_{j=1} f(r^*_{ij}, \theta^*_{ij}) r^*_{ij} \Delta r \Delta \theta $$

This becomes the expression for the double integral:

$$\iint\limits_R f(r,\theta) dA = \lim_{m,n \to \infty} \sum^m_{i=1} \sum^n_{j=1} f(r^*_{ij}, \theta^*_{ij}) r^*_{ij} \Delta r \Delta \theta $$

The double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates:

$$\iint\limits_R f(r,\theta) dA = \iint\limits_R f(r,\theta) r \ dr d\theta = \int_{\theta = \alpha}^{\theta = \beta} \int_{r=a}^{r=b} f(r,\theta) r \ dr d\theta$$

If \(f\) is given in terms of \(x\) and \(y\), then:

$$\iint\limits_R f(x,y) dA = \iint\limits_R f(r \cos \theta, r \sin \theta) r \ dr d\theta $$

Styles

(uses cookies)