21. Multiple Integrals in Curvilinear Coordinates
a. Integrating in Polar Coordinates
2. Integral over a Polar Rectangle
a. Derivation
We want to derive:
If \(R\) is the polar rectangle \([a,b]\times[\alpha,\beta]\), then \[ \iint\limits_R f(x,y)\,dA =\int_\alpha^\beta\int_a^b f(r,\theta)\,r\,dr\,d\theta =\int_a^b\int_\alpha^\beta f(r,\theta)\,r\,d\theta\,dr \] In particular, for a polar rectangle, the iterated integral is independent of the order of integration.
Add plots.Recall the Riemann sum definition of a double integral, and its implementation for a rectangle in rectangular coordinates. We now compute the double integral \(\displaystyle \iint\limits_R f(r,\theta)\,dA\) over a region \(R\) which is the polar rectangle, \(a \le r \le b\) and \(\alpha \le \theta \le \beta\).
Step 1 To partition \(R\) into subregions, first partition the \(r\) and \(\theta\) intervals into subintervals using partition points: \[\begin{aligned} r_0&=a \lt r_1 \lt r_2 \lt \cdots \lt r_{p-1} \lt r_p=b \\ \theta_0 &=\alpha \lt \theta_1 \lt \theta_2 \lt \cdots \lt \theta_{q-1} \lt \theta_q=\beta \end{aligned}\] Then the subregions are the polar rectangles: \[ R_{ij}=[r_{i-1},r_i]\times[\theta_{j-1},\theta_j] \] which have area: \[ \Delta A_{ij}=\bar{r}\Delta r_i\,\Delta\theta_j \] where the coordinate intervals are: \[ \Delta r_i=r_i-r_{i-1} \qquad \text{and} \qquad \Delta\theta_j=\theta_j-\theta_{j-1} \] and the average radius is: \[ \bar{r}=\dfrac{r_{i-1}+r_i}{2} \]
Step 2. Pick an evaluation point \(r_i^*\) in the subinterval \([r_{i-1},r_i]\) and an evaluation point \(\theta_j^*\) in the subinterval \([\theta_{j-1},\theta_j]\). Then \((r_i^*,\theta_j^*)\) is the evaluation point for the region \(R_{ij}\). In addition, approximate the average radius by the radius at the evaluation point in the area of a cell: \[ \Delta A_{ij}\approx r_i^*\Delta r_i\,\Delta\theta_j \]
Step 3. Evaluate the function \(f(r,\theta)\) at each evaluation point and multiply by the area of the polar cell. Add these up to get the Riemann sum: \[ \sum_{j=1}^q\sum_{i=1}^p f(r_i^*,\theta_j^*)\,r_i^*\,\Delta r_i\,\Delta\theta_j \]
Step 4. Then the double integral of \(f(r,\theta)\) over the region \(R\) is the limit of Riemann sums: \[ \iint\limits_R f(r,\theta)\,dA =\lim_{\begin{aligned}&\scriptstyle\quad q\rightarrow\infty \\ &\scriptstyle\text{max}\Delta\theta_j\rightarrow0\end{aligned}} \lim_{\begin{aligned}&\scriptstyle\quad p\rightarrow\infty \\ &\scriptstyle\text{max}\Delta r_i\rightarrow0\end{aligned}} \sum_{j=1}^q\sum_{i=1}^p f(r_i^*,\theta_j^*)\,r_i^*\,\Delta r_i\,\Delta\theta_j \]
Step 5a. Since the limit of a sum is the sum of the limits, and \(\Delta\theta_j\) is independent of the index \(i\) and the limit variable \(p\), we can write this as \[ \iint\limits_R f(r,\theta)\,dA =\lim_{\begin{aligned}&\scriptstyle\quad q\rightarrow\infty \\ &\scriptstyle\text{max}\Delta\theta_j\rightarrow0\end{aligned}} \sum_{j=1}^q \left( \lim_{\begin{aligned}&\scriptstyle\quad p\rightarrow\infty \\ &\scriptstyle\text{max}\Delta r_i\rightarrow0\end{aligned}} \sum_{i=1}^p f(r_i^*,\theta_j^*)\,r_i^*\,\Delta r_i \right)\Delta\theta_j \] We recognize this as the double iterated integral: \[ \iint\limits_R f(r,\theta)\,dA =\int_\alpha^\beta\int_a^b f(r,\theta)\,r\,dr\,d\theta \] where we first compute the \(r\) integral holding \(\theta\) fixed and then do the \(\theta\) integral.
Step 5b. The original double integral, in Step 4, can also be written as \[ \iint\limits_R f(r,\theta)\,dA =\lim_{\begin{aligned}&\scriptstyle\quad p\rightarrow\infty \\ &\scriptstyle\text{max}\Delta r_i\rightarrow0\end{aligned}} \lim_{\begin{aligned}&\scriptstyle\quad q\rightarrow\infty \\ &\scriptstyle\text{max}\Delta\theta_j\rightarrow0\end{aligned}} \sum_{i=1}^p\sum_{j=1}^q f(r_i^*,\theta_j^*)\,r_i^*\Delta\theta_j\,\Delta r_i\, \] by interchanging the order of the limits and sums. The same argument, as in Step 5a but interchanging \(r\) and \(\theta\), \(p\) and \(q\), and \(i\) and \(j\), gives \[ \iint\limits_R f(x,y)\,dA =\int_a^b\int_\alpha^\beta f(r,\theta)\,r\,d\theta\,dr \] where we first compute the \(\theta\) integral holding \(r\) fixed and then do the \(r\) integral.