While we have naturally defined double integrals in the rectangular coordinate system, starting with domains that are rectangular regions, there are many of these integrals that are difficult, if not impossible, to evaluate. For example, consider the domain \(D\) that is the unit circle (centered at the origin) and \(f(x,y) = e^{-x^2 - y^2}\text{.}\) To integrate \(f\) over \(D\text{,}\) we would use the iterated integral
For this particular integral, regardless of the order of integration, we are unable to find an antiderivative of the integrand; in addition, even if we were able to find an antiderivative, the inner limits of integration involve relatively complicated functions.
It is therefore useful to be able to translate to other coordinate systems where the limits of integration and the evaluation of the iterated integrals are simpler. In this section we provide a brief reminder of polar coordinates, then investigate their usage for double integrals. For a more thorough discussion of polar coordinates, look at Section 9.8.
The rectangular coordinate system is best suited for graphs and regions that are naturally considered over a rectangular grid. The polar coordinate system is an alternative that offers good options for functions and domains that have more circular characteristics. A point \(P\) in rectangular coordinates that is described by an ordered pair \((x,y)\text{,}\) where \(x\) is the displacement from \(P\) to the \(y\)-axis and \(y\) is the displacement from \(P\) to the \(x\)-axis can also be described with polar coordinates \((r,\theta)\text{,}\) where \(r\) is the distance from \(P\) to the origin and \(\theta\) is the angle formed by the line segment \(\overline{OP}\) and the positive \(x\)-axis, as shown at left in Figure 12.5.2.
Remember that you will need to interpret the results of \(r\) as being positive or negative along with interpreting the quadrant of \(\theta\text{.}\) You can look at Example 9.8.12 and later examples for different ways to interpret and make conversions between rectangular and polar coordinates.
We can draw graphs of curves in polar coordinates in a similar way to how we do in rectangular coordinates. However, when plotting in polar coordinates, we use a grid that considers changes in angles and changes in distance from the origin. In particular, the angles \(\theta\) and distances \(r\) partition the plane into small wedges as shown in Figure 12.5.2.
Draw a plot of each of the curves in the plane described by each of the following polar equations and write a sentence for each graph to explain why the equation is satisfied.
\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 where \(D\) has a vertically simple description (\(a\leq x \leq b, g_1(x) \leq y \leq g_2(x)\)). Note that the area element \(dA\) became \(dy\, dx \) in this conversion of the double integral into an iterated integral. We will need to transform the three different elements of this double integral into polar 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.
Remember that our classic calculus approach to defining the double integral involved breaking each coordinate into smaller pieces which created a collection of smaller areas that we used to create our Riemann sum. We can do the same procedure in terms of polar coordinates (instead of rectangular coordinates) and we could have our smaller pieces be in terms of some small changes in \(r\) and \(\theta\) (like the regions on the polar grid in Figure 12.5.2). We will now need to understand how to measure the area of a piece described by \(R_0\leq r \leq R_1 \quad \theta_0 \leq \theta \leq \theta_1\text{,}\) as shown in Figure 12.5.4.
We will first consider the area of the annulus between \(r=R_0\) and \(r=R-1\text{,}\) shown in Figure 12.5.3. The area of this region is the difference between the area of a disk of radius \(R_1\) and the area of a disk of radius \(R_0\text{.}\)
We want to measure the portion of the annulus that corresponds to \(\theta_0 \leq \theta \leq \theta_1\text{.}\) Let \(\Delta \theta=\theta_1-\theta_0\) be the change in the theta coordinate for the small piece we want to find the area of. Because we are using radian measures for our angles, the portion of the annulus we are using is \(\frac{\Delta \theta}{2\pi}\text{.}\) This means that the area of our small piece will be
\begin{equation*}
\Delta A = \frac{\Delta \theta}{2\pi} \left(\pi\left(R_1^2-R_0^2\right)\right)
\end{equation*}
Remember that we can easily convert our function \(f(x,y)\) to polar coordinates with our conversion equations to be \(f(r\cos(\theta),r\sin(\theta))\text{.}\) We can then approximate a double integral \(\iint_D f(x,y) \, dA\) with a Riemann sum of the form
We only have the second step of our conversion left above: convert the region of integration described by inequalities that correspond to fixing one coordinate and bounding the other in a consistent way. Similar to how we used vertically simple and horizontally simple descriptions for rectangular regions, we want to set up inequalities by hold one of our coordinates constant and setting boundary functions for the other coordinate.
Each of the types of regions above corresponds to exactly the necessary description to use a iterated integral. Let’s look at the inequalities and plot for a radially simple region; For any fixed value of \(\theta\) in the interval \([a,b]\text{,}\) we want to look at a slice along this constant \(\theta\) value. This corresponds to a ray from the origin in the cone between the grey lines in Figure 12.5.6. Along this \(\theta=k\) ray, the region of interest has an upper and lower bound on the \(r\)-coordinate. In fact, there is the function that describes the upper boundary, \(r=g_2(\theta)\text{,}\) for every value of \(\theta\) in \([a,b]\text{.}\) Similarly, there is the function that describes the lower boundary, \(r=g_1(\theta)\text{,}\) for every value of \(\theta\) in \([a,b]\text{.}\) Geometrically, this means that along every gray, dashed slice (where \(\theta\) is constant), there is the same upper and lower boundary curves, shown in red and blue, respectively.
While angularly simple regions are used much less often than radially simple regions, a parallel structure is used; Along every slice with a constant value of \(r\) in \([c,d]\) (the gray, dashed curves in Figure 12.5.7), the upper bound on \(\theta\) is given by the function \(\theta=h_2(r)\) and the lower bound is given by \(\theta=h_1(r)\) (shown by the red and blue bounding curves in Figure 12.5.7).
Just as in the rectangular coordinate case, especially complicated regions may need to be split into several different regions that are radially or angularly simple.
The double integral \(\iint_D f(x,y) \, dA\) in rectangular coordinates can be converted to a double integral in polar coordinates as \(\iint_D f(r\cos(\theta), r\sin(\theta)) \, r \, dr \, d\theta\text{.}\)
Note again that the order of integration corresponds to the manner in which we have sliced the region of integration, with the variable being held constant along a slice being the last variable to be integrated.
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 inequalities \(0 \leq r \leq 1\) and \(0 \leq \theta \leq 2\pi\text{.}\) Note that our region \(D\) can be consider both radially simple and angularly simple, with all of the bounds of integration being constant. In Figure 12.5.10, we show how \(D\) can be considered radially simple.
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 \(\sqrt{x^2 + y^2}\text{.}\)
Evaluate one of the iterated integrals. Write a couple sentences to explain your answer in the context of a double integral (as either a signed volume or an average value).
In the next activity, we will convert an equation from rectangular to polar coordinates and use this polar equation to give bounds of integration for two different integrals.
Determine a polar curve in the form \(r = f(\theta)\) that traces out the circle \(x^2 + (y-1)^2 = 1\text{.}\) You should substitute \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) into the rectangular coordinate equation and solve for \(r\text{.}\)
Find the volume under the surface \(h(x,y) = x\) over the region \(D\text{,}\) where \(D\) is the region bounded above by the line \(y=x\) and below by the circle (this is the shaded region in Figure 12.5.12).
To convert the double integral \({\iint_D f(x,y) \, dA}\) to an iterated integral in polar coordinates, we substitute \(r \cos(\theta)\) for \(x\text{,}\)\(r \sin(\theta)\) for \(y\text{,}\) and \(r \, dr \, d\theta\) for \(dA\) to obtain the iterated integral (where the order of integration is determined by whether the region is described as radially or angularly simple)
\begin{equation*}
{\iint_D f(r\cos(\theta), r\sin(\theta)) \, r \, dr \, d\theta}
\end{equation*}
For each of the following, set up the integral of an arbitrary function \(f(x,y)\) over the region in whichever of rectangular or polar coordinates is most appropriate. (Use \(t\) for \(\theta\) in your expressions.)
Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles \(x^2 + y^2 = 36\) and \(x^2 - 6x + y^2 = 0\text{.}\)
(a) Graph \(r=1/(4\cos\theta)\) for \(-\pi/2\le\theta\le\pi/2\) and \(r=1\text{.}\) Then write an iterated integral in polar coordinates representing the area inside the curve \(r=1\) and to the right of \(r=1/(4\cos\theta)\text{.}\) (Use \(t\) for \(\theta\) in your work.)
Using polar coordinates, evaluate the integral \(\displaystyle \int \!\! \int_{R} \sin (x^2+y^2) dA\) where R is the region \(4 \leq x^2 + y^2 \leq 25\text{.}\)
Instructions: Please enter the integrand in the first answer box, typing theta for \(\theta\text{.}\) Depending on the order of integration you choose, enter dr and dtheta in either order into the second and third answer boxes with only one dr or dtheta in each box. Then, enter the limits of integration.
Find the exact center of mass of the lamina over the portion of \(D\) that lies in the first quadrant and has its mass density distribution given by \(\delta(x,y) = 1\text{.}\) (Before making any calculations, where do you expect the center of mass to lie? Why?)