Evaluating Double Integrals in Polar Coordinates

Evaluating Double Integrals in Polar Coordinates

So we have looked at evaluating double integrals over general domains, however, sometimes it may be rather difficult to compute double integrals over certain domains due to the nature of integrating with the rectangular coordinate system. For example, computing a double integral over a circle with equation $(x - h)^2 + (y - k)^2 = r^2$.

Sometimes, it may be easier to use the polar coordinate system to compute a double integral. We will see that we can often times do this using the following formula to convert our double integral for a continuous function $f$ over the rectangle $R = \{ (x, y) : a ≤ r ≤ b, \alpha ≤ \theta ≤ \beta \}$ and where $0 ≤ \beta - \alpha ≤ 2 \pi$:

\begin{align} \quad \iint_R f(x, y) \: dA = \int_{\alpha}^{\beta} \int_a^b f(r \cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

Recall that if $(x, y)$ is a point in $\mathbb{R}^2$, then the relationship between $(x, y)$ and the corresponding polar coordinates $(r, \theta)$ come from the equations:

\begin{align} \quad x = r \cos \theta \quad , \quad y = r \sin \theta \end{align}

Furthermore, $r^2 = x^2 + y^2$. Sometimes it is easier to represent certain regions $R$ as $R = \{ (r, \theta) : a ≤ r ≤ b , \alpha ≤ \theta ≤ \beta \}$.


Now let's see how we would compute $\iint_R f(x, y) \: dA$. We start by taking the interval $[a, b]$ and dividing it into $m$ equal width subintervals $[ r_{i-1}, r_i ]$ where the width of each subinterval is $\Delta r = \frac{b - a}{m}$. We will also divide the interval $[ \alpha, \beta ]$ up into $n$ equal width subintervals $[\theta_{j-1}, \theta_j]$ where the width of each subinterval is $\Delta \theta = \frac{\beta - \alpha}{n}$. $R_{ij}$ will denote the polar sub-rectangle for which $r_{i-1} ≤ r ≤ r_{i}$ and $\theta_{j-1} ≤ \theta ≤ \theta_{j}$. The center of each polar subrectangle $R_{ij}$ which we will denote as $(r_i^*, \theta_j^*)$ is such that $r_i^* = \frac{1}{2} \left ( r_{i-1} + r_i \right )$ and $\theta_j = \frac{1}{2} \left ( \theta_{j-1}, \theta_j \right )$.


Now recall that the area of a sector of a circle is given by the formula:

\begin{align} \mathrm{Area \: of \: Sector} = \frac{1}{2} r^2 \theta \end{align}

Therefore we can compute the areas of each $R_{ij}$ as:

\begin{align} \quad \Delta A_i = \frac{1}{2} \left ( r_i^2 - r_{i-1}^2 \right ) \Delta \theta = \frac{1}{2} (r_i + r_{i-1})(r_i - r_{i-1}) \Delta \theta = r_i^* \Delta r \Delta \theta \end{align}

Using the conversion equations above, we have that the centers of each $R_{ij}$ in rectangular coordinates is $(r_i^* \cos \theta_j^*, r_i^* \sin \theta_j^*)$. Now the Riemann sum for setting up our double integral will be:

\begin{align} \quad \sum_{i=1}^{m} \sum_{j=1}^{n} f(r_i^* \cos \theta_j^*, r_i^* \sin \theta_j^*) \Delta A_i = \sum_{i=1}^{m} \sum_{j=1}^{n} f(r_i^* \cos \theta_j^*, r_i^* \sin \theta_j^*) r_i^* \Delta r \Delta \theta \end{align}

Now let $g(r, \theta) = f(r \cos \theta, r \sin \theta) \cdot r$ then we have that:

\begin{align} \sum_{i=1}^{m} \sum_{j=1}^{n} g(r_i^*, \theta_j^*) \Delta r \Delta \theta \end{align}

Thus we can evaluate our double integral as:

\begin{align} \quad \iint_R f(x, y) \: dA = \lim_{m, n \to \infty} \sum_{i=1}^{m} \sum_{j=1}^{n} f(r_i^* \cos \theta_j^*, r_i^* \sin \theta_j^*) \Delta A_i = \int_{\alpha}^{\beta} \int_a^b g(r, \theta) \: dr \: d \theta = \int_{\alpha}^{\beta} \int_a^b f(r \cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

Another type of region that we may need to integrate over is one such as $D = \{ (r, \theta) : h_1 (\theta) ≤ r ≤ h_2 (\theta) , \alpha ≤ \theta ≤ \beta \}$:


The double integral of $f(x, y)$ over $D$ can be computed as:

\begin{align} \quad \iint_D f(x, y) \: dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2 (\theta)} f(r \cos \theta, r \sin \theta) r \: dr \: d \theta \end{align}

What's nice is that we can also compute the area of polar region using double integrals.

Theorem 1: The area enclosed in a polar region $D$ bounded by $\alpha ≤ \theta ≤ \beta$ and $0 ≤ r ≤ h(\theta)$ is given by $\mathrm{Area} = \iint_D 1 \: dA$.
  • Proof: Suppose that $D$ is bounded by $\theta = \alpha$, $\theta = \beta$, $r = 0$, and $r = h(\theta)$. Then:
\begin{align} \quad \iint_ 1 \: dA = \int_{\alpha}^{\beta} \int_0^{h(\theta)} r \: dr \: d \theta = \int_{\alpha}^{\beta} \left [ \frac{r^2}{2} \right ]_{r=0}^{h(\theta)} \: d \theta = \frac{1}{2} \int_{\alpha}^{\beta} [h(\theta)]^2 \: d \theta \end{align}

Now the following Theorem gives us the value for a very important single integral that is widely used in areas such as probability.

Theorem 2: $\int_{-\infty}^{\infty} e^{-x^2} \: dx = \sqrt{\pi}$.
  • Proof: Let $I = \int_{-\infty}^{\infty} e^{-x^2} \: dx$. Note that we can also write $I = \int_{-\infty}^{\infty} e^{-y^2} \: dy$ as the choice of variable does not matter. Thus:
\begin{align} \quad I^2 = \left ( \int_{-\infty}^{\infty} e^{-x^2} \: dx \right ) \left ( \int_{-\infty}^{\infty} e^{-y^2} \: dy \right ) \\ \quad I^2 = \iint_{\mathbb{R}^2} e^{-x^2}e^{-y^2} \: dx \: dy \\ \quad I^2 = \iint_{\mathbb{R}^2} e^{-(x^2 + y^2)} \: dA \end{align}
  • We now express $\mathbb{R}^2$ as a polar region. We have that $\mathbb{R}^2 = \{ 0 ≤ r ≤ \infty, 0 ≤ \theta ≤ \pi \}$, and so:
\begin{align} \quad I^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r \: dr \: d \theta \\ \quad I^2 = \int_0^{2\pi} \lim_{b \to \infty} \left [ - \frac{1}{2} e^{-r^2} \right ]_{r=0}^{r=b} \: d \theta \\ \quad I^2 = \int_0^{2\pi} -\frac{1}{2} \left [-1 + \lim_{b \to \infty} e^{-b^2} \right ] \: d \theta \\ \quad I^2 = \int_0^{2\pi} \frac{1}{2} \: d \theta \\ \quad I^2 = \frac{1}{2} \left [ \theta \right ]_{\theta =0}^{\theta=2\pi} \\ \quad I^2 = \pi \\ \quad I = \sqrt{\pi} \quad \blacksquare \end{align}

Let's now look at some examples of calculating integrals with polar coordinates.

Example 1

Evaluate the double integral $\iint_R x + 2y \: dA$ where $R$ is the region of area between the semicircles above the $x$-axis of $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.

The region $R$ can be described as $R = \{ (r, \theta) : 1 ≤ r ≤ 2, 0 ≤ \theta ≤ \pi \}$:


We note that $f(x, y) = x + 2y$ and so $f(r \cos \theta, r \sin \theta) = r \cos \theta + 2r \sin \theta$. Using the formula above, we can convert our double integral into the following iterated integrals:

\begin{align} \quad \quad \iint_R x + 2y \: dA = \int_0^{\pi} \int_1^2 r^2 \cos \theta + 2r^2 \sin \theta \: dr \: d \theta = \int_0^{\pi} \left [ \frac{r^3}{3} \cos \theta + \frac{2r^3}{3} \sin \theta \right ]_{r=1}^{r=2} \: d \theta = \int_0^{\pi} \frac{8}{3} \cos \theta + \frac{16}{3} \sin \theta - \frac{1}{3} \cos \theta - \frac{2}{3} \sin \theta \: d \theta \\ = \int_0^{\pi} \frac{7}{3} \cos \theta + \frac{14}{3} \sin \theta \: d \theta = \left [ \frac{7}{3} \sin \theta - \frac{14}{3} \cos \theta \right ]_{\theta = 0}^{\theta = \pi} = \frac{28}{3} \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License