Consider the double integral
\begin{equation*}
\iint_D e^{x^2+y^2} \, dA,
\end{equation*}
where \(D\) is the unit disk. While we cannot directly evaluate this integral in rectangular coordinates, a change to polar coordinates will convert it to one we can easily evaluate.
We have seen how to evaluate a double integral \(\displaystyle \iint_D f(x,y) \, dA\) as an iterated integral of the form
\begin{equation*}
\int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx
\end{equation*}
in rectangular coordinates, because we know that \(dA = dy \, dx\) in rectangular coordinates. To make the change to polar coordinates, we not only need to represent the variables \(x\) and \(y\) in polar coordinates, but we also must understand how to write the area element, \(dA\text{,}\) in polar coordinates. That is, we must determine how the area element \(dA\) can be written in terms of \(dr\) and \(d\theta\) in the context of polar coordinates. We address this question in the following activity.
From the result of
Activity 11.5.3, we see when we convert an integral from rectangular coordinates to polar coordinates, we must not only convert
\(x\) and
\(y\) to being in terms of
\(r\) and
\(\theta\text{,}\) but we also have to change the area element to
\(dA = r \, dr \, d\theta\) in polar coordinates. As we saw in
Activity 11.5.3, the reason the additional factor of
\(r\) in the polar area element is due to the fact that in polar coordinates, the cross sectional area element increases as
\(r\) increases, while the cross sectional area element in rectangular coordinates is constant. So, given a double integral
\(\iint_D f(x,y) \, dA\) in rectangular coordinates, to write a corresponding iterated integral in polar coordinates, we replace
\(x\) with
\(r \cos(\theta)\text{,}\) \(y\) with
\(r \sin(\theta)\) and
\(dA\) with
\(r \, dr \, d\theta\text{.}\) Of course, we need to describe the region
\(D\) in polar coordinates as well. To summarize:
Example 11.5.3.
Let
\(f(x,y) = e^{x^2+y^2}\) on the disk
\(D = \{(x,y) : x^2 + y^2 \leq 1\}\text{.}\) We will evaluate
\(\iint_D f(x,y) \, dA\text{.}\)
In rectangular coordinates the double integral \(\iint_D f(x,y) \, dA\) can be written as the iterated integral
\begin{equation*}
\iint_D f(x,y) \, dA = \int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} e^{x^2+y^2} \, dy \, dx.
\end{equation*}
We cannot evaluate this iterated integral, because \(e^{x^2 + y^2}\) does not have an elementary antiderivative with respect to either \(x\) or \(y\text{.}\) However, since \(r^2=x^2+y^2\) and the region \(D\) is circular, it is natural to wonder whether converting to polar coordinates will allow us to evaluate the new integral. To do so, we replace \(x\) with \(r \cos(\theta)\text{,}\) \(y\) with \(r \sin(\theta)\text{,}\) and \(dy \, dx\) with \(r \, dr \, d\theta\) to obtain
\begin{equation*}
\iint_D f(x,y) \, dA = \iint_D e^{r^2} \, r \, dr \, d\theta.
\end{equation*}
The disc \(D\) is described in polar coordinates by the constraints \(0 \leq r \leq 1\) and \(0 \leq \theta \leq 2\pi\text{.}\) Therefore, it follows that
\begin{equation*}
\iint_D e^{r^2} \, r \, dr \, d\theta = \int_{\theta=0}^{\theta = 2\pi} \int_{r=0}^{r=1} e^{r^2} \, r \, dr \, d\theta.
\end{equation*}
We can evaluate the resulting iterated polar integral as follows:
\begin{align*}
\int_{\theta=0}^{\theta = 2\pi} \int_{r=0}^{r=1} e^{r^2} \, r \, dr \, d\theta \amp = \int_{\theta=0}^{2\pi} \left( \frac{1}{2}e^{r^2}\biggm|_{r=0}^{r=1} \right) \, d\theta\\
\amp = \frac{1}{2} \int_{\theta=0}^{\theta = 2\pi} \left( e-1 \right) \, d\theta\\
\amp = \frac{1}{2}(e-1) \int_{\theta=0}^{\theta = 2\pi} \, d\theta\\
\amp = \frac{1}{2}(e-1)\left[\theta\right]\biggm|_{\theta=0}^{\theta = 2\pi}\\
\amp = \pi(e-1).
\end{align*}
While there is no firm rule for when polar coordinates can or should be used, they are a natural alternative anytime the domain of integration may be expressed simply in polar form, and/or when the integrand involves expressions such as
\(\sqrt{x^2 + y^2}.\)
