3 Evaluating surface integrals using polar coordinates

Areas with circular boundaries often lead to double integrals with awkward limits, and these integrals can be difficult to evaluate. In such cases it is easier to work with polar $\left(r,\theta \right)$ rather than Cartesian $\left(x,y\right)$ coordinates.

3.1 Polar coordinates

Figure 17

The polar coordinates of the point $P$ are the distance $r$ from $P$ to the origin $O$ and the angle $\theta$ that the line $OP$ makes with the positive $x$ axis. The following are used to transform between polar and rectangular coordinates.

1. Given $\left(x,y\right)$ , $\left(r,\theta \right)$ are found using $r=\sqrt{x^2+y^2}$ and $tan\theta =\frac{y}{x}$ .
2. Given $\left(r,\theta \right)$ , $\left(x,y\right)$ are found using $x=rcos\theta$ and $y=rsin\theta$

Note that we also have the relation $r^2=x^2+y^2$ .

3.2 Finding surface integrals with polar coordinates

The area of integration $A$ is covered with coordinate circles given by $r=\text{constant}$ and coordinate lines given by $\theta =\text{constant}$ .

The elementary areas $\delta A$ are almost rectangles having width $\delta r$ and length determined by the length of the part of the circle of radius $r$ between $\theta$ and $\delta \theta$ , the arc length of this part of the circle is $r\delta \theta$ . So $\delta A\approx r\delta r\delta \theta$ . Thus to evaluate ${\int \; <!--nolimits\; -->}_{A}f\left(x,y\right)dA$ we sum $f\left(r,\theta \right)r\delta r\delta \theta$ for all $\delta A$ .

${\int \; <!--nolimits\; -->}_{A}f\left(x,y\right)dA={\int \; <!--nolimits\; -->}_{\theta ={\theta }_A}^{\theta ={\theta }_B}{\int \; <!--nolimits\; -->}_{r=r_1\left(\theta \right)}^{r=r_2\left(\theta \right)}f\left(r,\theta \right)rdrd\theta$

Figure 18

Note that the use of polar coordinates is a special case of the use of a change of variables. Further cases of change of variables will be considered in Section 27.4.

Example 14

Evaluate ${\int \; <!--nolimits\; -->}_0^{\frac{\pi }{3}}{\int \; <!--nolimits\; -->}_0^{2}rcos\theta drd\theta$ and sketch the region of integration. Note that it is the function $cos\theta$ which is being integrated over the region and the $r$ comes from the $rdrd\theta$ .

Figure 19

Solution

The evaluation is similar to that for cartesian coordinates. The inner integral with respect to $r$ , is evaluated first with $\theta$ constant. Then the outer $\theta$ integral is evaluated.

With $\theta$ constant $r$ varies between $0$ and $2$ , so the bounding curves of the polar strip start at $r=0$ and end at $r=2$ . As $\theta$ varies between $0$ and $\frac{\pi }{3}$ a sector of a circular disc is swept out. This sector is the region of integration shown above.

Example 15

Earlier in this Section, an example concerned integrating the function $f\left(x,y\right)=5x^{2}y$ over the half of the unit circle which lies above the $x$ -axis. It is also possible to carry out this integration using polar coordinates.

Solution

The semi-circle is characterised by $0\le r\le 1$ and $0\le \theta \le \pi$ . So the integral may be written (remembering that $x=rcos\theta$ and $y=rsin\theta$ )

${\int \; <!--nolimits\; -->}_0^{\pi }{\int \; <!--nolimits\; -->}_0^{1}5{\left(rcos\theta \right)}^2\left(rsin\theta \right)rdrd\theta$

which can be evaluated as follows

This is, of course, the same answer that was obtained using an integration over rectangular coordinates.