21. Multiple Integrals in Curvilinear Coordinates

a. Integrating in Polar Coordinates

3. Integral over a Non-Rectangular Polar Region

a. Derivation

We start with regions of Type \(r\):

If \(R\) is a Type \(r\) region with \(a \le r \le b\) and \(g(r) \le \theta \le h(r)\), then \[ \iint\limits_R f(r,\theta)\,dA =\int_a^b\int_{g(r)}^{h(r)} f(r,\theta)\,r\,d\theta\,dr \] This is just the iterated integral computed by holding \(r\) fixed while doing the \(\theta\) integral and then integrating the resulting function over \(r\).

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We want to compute \(\displaystyle \iint\limits_R f(r,\theta)\,dA\) over a Type \(r\) region \(R\) satisfying \(a \le r \le b\) and \(g(r) \le \theta \le h(r)\). We partition the rectangle \(R\) into subregions in two steps.

Step 1. First partition the \(r\) interval into subintervals using the partition points: \[ r_0=a \lt r_1 \lt r_2 \lt \cdots \lt r_{p-1} \lt r_p=b \] and pick an evaluation point \(r_i^*\) in each subinterval \([r_{i-1},r_i]\).

Step 2. Then for each subinterval \([r_{i-1},r_i]\), partition the \(\theta\) interval \([g(r_i^*),h(r_i^*)]\) using the partition points \[ \theta_{i,0} =g(r_i^*) \lt \theta_{i,1} \lt \theta_{i,2} \lt \cdots \lt \theta_{i,q-1} \lt \theta_{i,q} =h(r_i^*) \] and pick an evaluation point \(\theta_{ij}^*\) in the subinterval \([\theta_{i,j-1},\theta_{i,j}]\).

Then the subregions are the polar rectangles \(R_{ij}=[r_{i-1},r_i]\times[\theta_{i,j-1},\theta_{i,j}]\) which have area \(\Delta A_{ij} =\bar r_i\Delta r_i\Delta\theta_{ij} =\dfrac{r_i+r_{i-1}}{2}(r_i-r_{i-1})(\theta_{i,j}-\theta_{i,j-1})\), and evaluation points \((r_i^*,\theta_{i,j}^*)\).

Notice that all the values of \(\theta\) carry the double index \(ij\) instead of the single index \(j\) because these values depend on the \(r\) subinterval \([r_{i-1},r_i]\) since the endpoints \(g(r_i^*)\) and \(h(r_i^*)\) are not constant; they depend on \(r_i^*\).

Step 3. Then the double integral of \(f(r,\theta)\) over the region \(R\) is the limit of Riemann sums: \[ \iint\limits_R f(x,y)\,dA =\lim_{\begin{aligned}&\scriptstyle\quad p\rightarrow\infty \\ &\scriptstyle\text{max}\Delta r_i\rightarrow0\end{aligned}} \lim_{\begin{aligned}&\scriptstyle\quad q\rightarrow\infty \\ &\scriptstyle\text{max}\Delta \theta_{ij}\rightarrow0\end{aligned}} \,\sum_{i=1}^p\sum_{j=1}^q f(r_i^*,\theta_{i,j}^*)\,r_i^*\,\Delta r_i\,\Delta\theta_{ij} \] Since the limit of a sum is the sum of the limits, we can write this as \[ \iint\limits_R f(r,\theta)\,dA =\lim_{\begin{aligned}&\scriptstyle\quad p\rightarrow\infty \\ &\scriptstyle\text{max}\Delta r_i\rightarrow0\end{aligned}} \,\sum_{i=1}^p \left( \lim_{\begin{aligned}&\scriptstyle\quad q\rightarrow\infty \\ &\scriptstyle\text{max}\Delta \theta_{ij}\rightarrow0\end{aligned}} \,\sum\limits_{j=1}^q f(r_i^*,\theta_{i,j}^*)\Delta\theta_{ij} \right)\,r_i^*\,\Delta r_i \] Notice that we have also factored the \(r_i^*\,\Delta r_i\) out of the sum \(j=1,\ldots,q\) because it is the same for all terms and also pulled it out of the limit \(q\rightarrow\infty\) since it is independent of \(q\). Now the quantity in the parentheses can be recognized as the single \(\theta\)-integral \(\displaystyle \int_{g(r_i^*)}^{h(r_i^*)} f(r_i^*,\theta)\,d\theta\). So \[ \iint\limits_R f(r,\theta)\,dA =\lim_{\begin{aligned}&\scriptstyle\quad p\rightarrow\infty \\ &\scriptstyle\text{max}\Delta r_i\rightarrow0\end{aligned}} \sum_{i=1}^p \left( \int_{g(r_i^*)}^{h(r_i^*)} f(r_i^*,\theta)\,d\theta \right)r_i^*\Delta r_i \] Finally, since \(\displaystyle \int_{g(r_i^*)}^{h(r_i^*)} f(r_i^*,\theta)\,d\theta\,r_i^*\) is itself a function of \(r\) evaluated at \(r=r_i^*\), the remaining limit of a sum is also an integral. So \[ \iint\limits_R f(r,\theta)\,dA =\int_a^b\int_{g(r)}^{h(r)} f(r,\theta)\,r\,d\theta\,dr \]

This is just the iterated integral computed by holding \(r\) fixed while doing the \(\theta\) integral and then integrating the resulting function over \(r\).

The main difference between this derivation and that for a polar rectangle is that the order of integration cannot be reversed because \(\Delta\theta_{ij}\) cannot be factored out of the sum \(i=1,\ldots,p\) because it depends on \(i\).

We now turn to regions of Type \(\theta\):

If \(R\) is a Type \(\theta\) region with \(\alpha \le \theta \le \beta\) and \(g(\theta) \le r \le h(\theta)\), then \[ \iint\limits_R f(r,\theta)\,dA =\int_\alpha^\beta\int_{g(\theta)}^{h(\theta)} f(r,\theta)\,r\,dr\,d\theta \] This is just the iterated integral computed by holding \(\theta\) fixed while doing the \(r\) integral and then integrating the resulting function over \(\theta\).

For a Type \(\theta\) region, the above derivation can be repeated with the rolls of \(r\) and \(\theta\) interchanged.