# Section13.3Double Integration with Polar Coordinates¶ permalink

We have used iterated integrals to evaluate double integrals, which give the signed volume under a surface, $z=f(x,y)\text{,}$ over a region $R$ of the $x$-$y$ plane. The integrand is simply $f(x,y)\text{,}$ and the bounds of the integrals are determined by the region $R\text{.}$

Some regions $R$ are easy to describe using rectangular coordinates — that is, with equations of the form $y=f(x)\text{,}$ $x=a\text{,}$ etc. However, some regions are easier to handle if we represent their boundaries with polar equations of the form $r=f(\theta)\text{,}$ $\theta = \alpha\text{,}$ etc.

The basic form of the double integral is $\iint_R f(x,y)\ dA\text{.}$ We interpret this integral as follows: over the region $R\text{,}$ sum up lots of products of heights (given by $f(x_i,y_i)$) and areas (given by $\Delta A_i$). That is, $dA$ represents “a little bit of area.” In rectangular coordinates, we can describe a small rectangle as having area $dx\ dy$ or $dy\ dx$ — the area of a rectangle is simply length×width — a small change in $x$ times a small change in $y\text{.}$ Thus we replace $dA$ in the double integral with $dx\ dy$ or $dy\ dx\text{.}$

Now consider representing a region $R$ with polar coordinates. Consider Figure 13.3.1(a). Let $R$ be the region in the first quadrant bounded by the curve. We can approximate this region using the natural shape of polar coordinates: portions of sectors of circles. In the figure, one such region is shaded, shown again in part (b) of the figure.

As the area of a sector of a circle with radius $r\text{,}$ subtended by an angle $\theta\text{,}$ is $A = \frac12r^2\theta\text{,}$ we can find the area of the shaded region. The whole sector has area $\frac12r_2^2\Delta \theta\text{,}$ whereas the smaller, unshaded sector has area $\frac12r_1^2\Delta \theta\text{.}$ The area of the shaded region is the difference of these areas: \begin{equation*} \Delta A_i = \frac12r_2^2\Delta\theta-\frac12r_1^2\Delta\theta = \frac12\big(r_2^2-r_1^2\big)\big(\Delta\theta\big) = \frac{r_2+r_1}{2}\big(r_2-r_1\big)\Delta\theta. \end{equation*}

Note that $(r_2+r_1)/2$ is just the average of the two radii.

To approximate the region $R\text{,}$ we use many such subregions; doing so shrinks the difference $r_2-r_1$ between radii to 0 and shrinks the change in angle $\Delta \theta$ also to 0. We represent these infinitesimal changes in radius and angle as $dr$ and $d\theta\text{,}$ respectively. Finally, as $dr$ is small, $r_2\approx r_1\text{,}$ and so $(r_2+r_1)/2\approx r_1\text{.}$ Thus, when $dr$ and $d\theta$ are small, \begin{equation*} \Delta A_i \approx r_i\ dr\ d\theta. \end{equation*}

Taking a limit, where the number of subregions goes to infinity and both $r_2-r_1$ and $\Delta\theta$ go to 0, we get \begin{equation*} dA = r\ dr\ d\theta. \end{equation*}

So to evaluate $\iint_Rf(x,y)\ dA\text{,}$ replace $dA$ with $r\ dr\ d\theta\text{.}$ Convert the function $z=f(x,y)$ to a function with polar coordinates with the substitutions $x=r\cos(\theta)\text{,}$ $y=r\sin(\theta)\text{.}$ Finally, find bounds $g_1(\theta)\leq r\leq g_2(\theta)$ and $\alpha\leq\theta\leq\beta$ that describe $R\text{.}$ This is the key principle of this section, so we restate it here as a Key Idea.

##### Key Idea13.3.4Evaluating Double Integrals with Polar Coordinates

Let $R$ be a plane region bounded by the polar equations $\alpha\leq\theta\leq\beta$ and $g_1(\theta)\leq r\leq g_2(\theta)\text{.}$ Then \begin{equation*} \iint_Rf(x,y)\ dA = \int_\alpha^\beta\int_{g_1(\theta)}^{g_2(\theta)} f\big(r\cos(\theta) ,r\sin(\theta) \big)\ r\ dr\ d\theta. \end{equation*}

Examples will help us understand this Key Idea.

##### Example13.3.5Evaluating a double integral with polar coordinates

Find the signed volume under the plane $z= 4-x-2y$ over the circle with equation $x^2+y^2=1\text{.}$

Solution
##### Example13.3.7Evaluating a double integral with polar coordinates

Find the volume under the paraboloid $z=4-(x-2)^2-y^2$ over the region bounded by the circles $(x-1)^2+y^2=1$ and $(x-2)^2+y^2=4\text{.}$

Solution
##### Example13.3.11Evaluating a double integral with polar coordinates

Find the volume under the surface $\ds f(x,y) =\frac1{x^2+y^2+1}$ over the sector of the circle with radius $a$ centered at the origin in the first quadrant, as shown in Figure 13.3.12.

Solution
##### Example13.3.13Finding the volume of a sphere

Find the volume of a sphere with radius $a\text{.}$

Solution
##### Example13.3.14Finding the volume of a solid

A sculptor wants to make a solid bronze cast of the solid shown in Figure 13.3.15, where the base of the solid has boundary, in polar coordinates, $r=\cos(3\theta)\text{,}$ and the top is defined by the plane $z=1-x+0.1y\text{.}$ Find the volume of the solid.

Solution

We have used iterated integrals to find areas of plane regions and volumes under surfaces. Just as a single integral can be used to compute much more than “area under the curve,” iterated integrals can be used to compute much more than we have thus far seen. The next two sections show two, among many, applications of iterated integrals.

# Subsection13.3.1Exercises

Terms and Concepts

In the following exercises, a function $f(x,y)$ is given and a region $R$ of the $x$-$y$ plane is described. Set up and evaluate $\iint_Rf(x,y)\ dA$ using polar coordinates.

In the following exercises, an iterated integral in rectangular coordinates is given. Rewrite the integral using polar coordinates and evaluate the new double integral.

In the following exercises, special double integrals are presented that are especially well suited for evaluation in polar coordinates.