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 \(r\).

The plot shows a bent rectangle with 2 edges which are arcs of
      circles labeled r = a and r = b. The other 2 edges are curves labeled
      theta = g of r and theta = h of r.
Type \(r\) Region

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.

First partition the \(r\) interval into subintervals using the partition points: \[ r_0=a \lt r_1 \lt r_2 \lt \cdots \lt r_{p-1} \lt r_p=b \]

The plot shows the polar rectangle of the previous plot, broken
      up into 4 strips by 3 circular arcs.

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 \([\theta_{i,j-1},\theta_{i,j}]\).

The plot shows the polar rectangle broken up into 4 strips as in
      the previous plot, but now each strip is broken into 4 smaller polar
      rectangles by 3 radial segments for a total of 16 boxes.

Pick an evaluation point \(r_i^*\) in each subinterval \([r_{i-1},r_i]\). Then for each subinterval \([r_{i-1},r_i]\), pick an evaluation point \(\theta_{ij}^*\) in the subinterval \([\theta_{i,j-1},\theta_{i,j}]\). Then the evaluation point in the \(ij\) rectangle is \((r_i^*,\theta_{ij}^*)\). They line up in the \(r\) direction but not in the \(\theta\) direction, because \(\theta_{ij}^*\) depends on \(i\) as well as \(j\).

The plot shows the polar rectangle broken up into 16 boxes as in
      the previous plot, but now each box contains a dot. The dots line up
      in the radial direction but not the angular direction.

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: \[ \Delta A_{ij} =\bar r_i\Delta r_i\Delta\theta_{ij} =\dfrac{r_i+r_{i-1}}{2}(r_i-r_{i-1})(\theta_{i,j}-\theta_{i,j-1}) \]

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 \(r_i^*\).

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 \(\theta\).

The plot shows a bent rectangle with 2 edges which are pieces of
    radial lines labeled theta = alpha and theta = beta. The other 2 edges
    are curves labeled r = g of theta and r = h of theta.
Type \(\theta\) Region

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

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Supported in part by NSF Grant #1123255