Double integrals in polar coordinates (article) | Khan Academy (2024)

• Polar coordinates (video)
• Double integrals beyond volume

• When you are performing a double integral,

  ${\displaystyle {\iint }_{R}fdA}$‍

  if you wish to express the function $f$‍ and the bounds for the region $R$‍ in polar coordinates $(r,\theta )$‍, the way to expand the tiny area $dA$‍ is

  $dA=rd\theta dr$‍

  (Pay attention to the fact that the variable $r$‍ is part of this expression)

• Beyond that one rule, these double integrals are mostly about being careful to make sure the bounds of your integrals appropriately encode the region $R$‍.

• Integrating using polar coordinates is handy whenever your function or your region have some kind of rotational symmetry. For example, polar coordinates are well-suited for integration in a disk, or for functions including the expression $x^2+y^2$‍.

Suppose we have a multivariable function defined using the polar coordinates $r$‍ and $\theta$‍,

$f(r,\theta )=r^2$‍

And let's say you want to find the double integral of this function in the region where

$r\le 2$‍

This is a disc of radius $2$‍ centered at the origin.

Written abstractly, here's what this double integral might look like:

$\begin{array}{r}{\iint }_{r\le 2}{r}^{2}dA\end{array}$‍

You could interpret this as the volume underneath a paraboloid (the three-dimensional analog of a parabola), as pictured below:

The question is, what do we do with that $dA$‍ term?

Warning!: You might be tempted to replace $dA$‍ with $d\theta dr$‍, since in cartesian coordinates we replace it with $dxdy$‍. But this is not correct!

Remember what a double integral is doing: It chops up the region that we are integrating over into tiny pieces, and $dA$‍ represents the area of each one of those pieces. For example, chopping up our disc of radius $2$‍ might look like this:

Why did I choose to chop it in this spiderweb pattern, as opposed to using vertical and horizontal lines? Since we are in polar coordinates, it will be easiest to think about the tiny pieces if their edges represent either a constant $r$‍ value or a constant $\theta$‍ value.

Let's focus on just one of these little chunks:

Even though this little piece has a curve shape, if we make finer and finer cuts, we can basically treat it as a rectangle. The length of one side of this "rectangle" can be thought of as $dr$‍, a tiny change in the $r$‍-coordinate.

Using a differential $dr$‍ to describe this length emphasizes the fact that we are not really considering a specific piece, but instead we care about what happens as its size approaches $0$‍.

But how long is the other side?

It's not $d\theta$‍, a tiny change in the angle, because radians are not a unit of length. To turn radians into a bit of arc length, we must multiply by $r$‍.

Therefore, if we treat this tiny chunk as a rectangle, and as $dr$‍ and $d\theta$‍ each approach $0$‍ it basically is a rectangle, its area is

$dA=(rd\theta )(dr)$‍

Plugging this into our original integral, we get

$\begin{array}{r}{\iint }_{r\le 2}{r}^{2}dA={\iint }_{r\le 2}{r}^2(rd\theta )(dr)={\iint }_{r\le 2}{r}^{3}d\theta dr\end{array}$‍

Putting bounds on this region is relatively straight-forward in this example, because circles are naturally suited for polar coordinates. Since we wrote $d\theta$‍ in front of $dr$‍, the inner integral is written with respect to $\theta$‍. The bounds of this inner integral will reflect the full range of $\theta$‍ as it sweeps once around the circle, going from $0$‍ to $2\pi$‍. The outer integral is with respect to $r$‍, which ranges from $0$‍ to $2$‍.

Concept check: Evaluate this double integral

$\begin{array}{r}{\int }_0^2{\int }_0^{2\pi }{r}^{3}d\theta dr=\end{array}$‍

$\begin{array}{rl}& {\int }_0^2\underset{\begin{array}{c}{r}^3\text{can be factored}\\ \text{out of the inner integral}\\ \text{with respect to}\theta \end{array}}{\underbrace{\left({\int }_0^{2\pi }{r}^{3}d\theta \right)}}dr\\ \\ & ={\int }_0^2{r}^3\left({\int }_0^{2\pi }d\theta \right)dr\\ \\ & ={\int }_0^2{r}^3(2\pi )dr\\ \\ & =2\pi {\int }_0^2{r}^{3}dr\\ \\ & =2\pi {\left({\displaystyle \frac{r^4}{4}}\right)}_0^2\\ \\ & =2\pi ({\displaystyle \frac{2^4}{4}}-{\displaystyle \frac{0^4}{4}})\\ \\ & =8\pi \end{array}$‍

Define a two-variable function $f$‍ in polar coordinates as

$f(r,\theta )=r\sin (\theta )$‍

Let $R$‍ be flower-shaped region, defined by

$r\le \cos (2\theta )$‍

Solve the double integral

$\begin{array}{r}{\iint }_{R}fdA\end{array}$‍

Step 1: Which of the following represents the right way to replace $fdA$‍ in the abstractly written double integral?

Choose 1 answer:

Choose 1 answer:

• (Choice A)

  $\begin{array}{r}{\iint }_{R}r\sin (\theta )drd\theta \end{array}$‍

• (Choice B)

  $\begin{array}{r}{\iint }_R{r}^2\sin (\theta )drd\theta \end{array}$‍

The second choice is correct:

$\begin{array}{r}{\iint }_R{r}^2\sin (\theta )drd\theta \end{array}$‍

The function $f$‍ is defined as $r\sin (\theta )$‍, and the term $dA$‍ is expanded as $rd\theta dr$‍ (don't forget to add the $r$‍ term here)

$\begin{array}{rl}& {\iint }_{R}fdA\\ \\ & ={\iint }_R\underset{f(r,\theta )}{\underbrace{r\sin (\theta )}}\underset{dA}{\underbrace{rd\theta dr}}\\ \\ & ={\iint }_R{r}^2\sin (\theta )drd\theta \end{array}$‍

Step 2: Now we must encode the fact that $R$‍ is defined as the region where $r\le \cos (2\theta )$‍. Which of the following is the right way to put bounds on the double integral?

Choose 1 answer:

Choose 1 answer:

• (Choice A)

  $\begin{array}{r}{\int }_0^{2\pi }{\int }_0^{\cos (2\theta )}{r}^2\sin (\theta )drd\theta \end{array}$‍

• (Choice B)

  $\begin{array}{r}{\int }_0^{\arccos (r)/2}{\int }_0^{2\pi }{r}^2\sin (\theta )drd\theta \end{array}$‍

The first answer choice is correct

$\begin{array}{r}{\int }_0^{2\pi }{\int }_0^{\cos (2\theta )}{r}^2\sin (\theta )drd\theta \end{array}$‍

The inner integral reflects $r$‍ bounds, since $dr$‍ is written before $d\theta$‍. Our region is explicitly defined such that $r\le \cos (2\theta )$‍, and since these are polar coordinates, $r$‍ can only ever be positive. Therefore, $r$‍ ranges from $0$‍ to $\cos (2\theta )$‍.

For the outer integral, $\theta$‍ ranges over its full range from $0$‍ to $2\pi$‍, since no constraints are given on $\theta$‍ in the definition of $R$‍.

Step 3: Solve this integral.

$\begin{array}{r}{\int }_0^{2\pi }{\int }_0^{\cos (2\theta )}{r}^2\sin (\theta )drd\theta =\end{array}$‍

In practice, I would use a calculator or Wolfram Alpha to solve one of these integrals. The hard part, which typically requires more human intelligence, is getting the double integral to the point where all the terms are in place. If you want to solve it by hand, here's how it might go:

$\begin{array}{rl}& {\int }_0^{2\pi }\underset{\text{Factor out}\sin (\theta )\text{from}r\text{-integral}}{\underbrace{{\int }_0^{\cos (2\theta )}{r}^2\sin (\theta )dr}}d\theta \\ \\ & ={\int }_0^{2\pi }\sin (\theta ){\int }_0^{\cos (2\theta )}{r}^{2}drd\theta \\ \\ & ={\int }_0^{2\pi }\sin (\theta ){\left[{\displaystyle \frac{r^3}{3}}\right]}_0^{\cos (2\theta )}d\theta \\ \\ & ={\int }_0^{2\pi }\sin (\theta )\left({\displaystyle \frac{\cos (2\theta {)}^3}{3}}\right)d\theta \\ \\ & ={\displaystyle \frac{1}{3}}{\int }_0^{2\pi }\sin (\theta )\cos (2\theta {)}^{3}d\theta \end{array}$‍

From here, we can save a lot of computational trouble with an observation: The inside of the integral is an odd function, meaning that when you replace $\theta$‍ by $-\theta$‍, the value of the expression as a whole becomes negative.

Also, since these trigonometric functions are periodic, integrating between $0$‍ and $2\pi$‍ is the same as integrating between $-\pi$‍ and $\pi$‍. The integral on the negative half cancels out with the integral on the positive portion, so in total it is $0$‍.

Are you ready for one of my favorite results in math? This is really quite clever.

Question: What is the integral $\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx\end{array}$‍ ?

This single integral is hard-to-impossible to compute directly. Just try to find the antiderivative!

This integral is asking about the area under a bell curve, which turns out to be super important for probability and statistics!

"What does this have to do with double integrals in polar coordinates?"

I hear you, my inquisitive friend, it does seem unrelated, doesn't it? Well, this is where someone got super clever.

Surprisingly, it is easier to solve this multi-dimensional analog of this problem. Namely, find the volume under a three-dimensional bell curve over the entire $xy$‍-plane.

$\begin{array}{r}{\iint }_{xy\text{-plane}}{e}^{-(x^2+y^2)}dA\end{array}$‍

If we keep everything in cartesian coordinates, this is as hard to solve as the original single integral.

$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dxdy\end{array}$‍

However, something magical happens when we convert to polar coordinates.

Concept check: Express this double integral using polar coordinates.

Choose 1 answer:

Choose 1 answer:

• (Choice A)

  $\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-r^2}rd\theta dr\end{array}$‍

• (Choice B)

  $\begin{array}{r}{\int }_0^{\mathrm{\infty }}{\int }_0^{2\pi }{e}^{-r^2}rd\theta dr\end{array}$‍

• (Choice C)

  $\begin{array}{r}{\int }_0^{\mathrm{\infty }}{\int }_0^{2\pi }{e}^{-r^2}d\theta dr\end{array}$‍

The second answer choice is correct.

First, make the following two conversions

$x^2+y^2\to r^2$‍

and

$\begin{array}{r}dydx\to \underset{\text{Polar form of}dA}{\underbrace{rd\theta dr}}\end{array}$‍

So our integral looks like this:

$\begin{array}{r}{\iint }_{r\theta \text{-plane}}{e}^{-r^2}rd\theta dr\end{array}$‍

To put bounds on the integral, we cover the entire plane by letting $\theta$‍ range from $0$‍ to $2\pi$‍, and $r$‍ range from $0$‍ to $\mathrm{\infty }$‍.

$\begin{array}{r}{\int }_0^{\mathrm{\infty }}{\int }_0^{2\pi }{e}^{-r^2}rd\theta dr\end{array}$‍

Since the inner integral is with respect to $\theta$‍, we can factor out everything with an $r$‍ in it, which in this case is the entire function:

$\begin{array}{rl}& {\int }_0^{\mathrm{\infty }}{\int }_0^{2\pi }{e}^{-r^2}rd\theta dr\\ \\ & ={\int }_0^{\mathrm{\infty }}{e}^{-r^2}r\underset{\text{This evaluates to}2\pi }{\underbrace{{\int }_0^{2\pi }d\theta }}dr\\ \\ & ={\int }_0^{\mathrm{\infty }}\left(e^{-r^2}r\right)(2\pi )dr\\ \\ & =2\pi {\int }_0^{\mathrm{\infty }}{e}^{-r^2}rdr\end{array}$‍

Concept check: Find the antiderivative of $e^{-r^2}r$‍, using either $u$‍-substitution or the inverse chain rule.

$\begin{array}{r}\int e^{-r^2}rdr=\end{array}$‍

Using u‍-substitution:

Define $u$‍ as

$u=-r^2$‍

which means

$du=-2rdr$‍

Our indefinite integral now becomes

$\begin{array}{rl}\int e^{-r^2}rdr& =\int e^{u}r{\displaystyle \frac{du}{-2r}}\\ \\ & =\int e^u{\displaystyle \frac{-1}{2}}du\\ \\ & =-{\displaystyle \frac{1}{2}}{e}^u\end{array}$‍

Substituting back to $u=-r^2$‍, this means our final antiderivative is

$\begin{array}{rl}\int e^{-r^2}rdr& =-{\displaystyle \frac{1}{2}}{e}^{-r^2}\end{array}$‍

Notice, the reason you can now find an antiderivative is because of that little $r$‍ term that showed up due to the fact that $dA=rd\theta dr$‍.

Concept check: Using this antiderivative, finish solving the integral which computes volume under a three-dimensional bell curve.

$\begin{array}{r}2\pi {\int }_0^{\mathrm{\infty }}{e}^{-r^2}rdr=\end{array}$‍

$\begin{array}{rl}2\pi {\int }_0^{\mathrm{\infty }}{e}^{-r^2}rdr& =2\pi {[-{\displaystyle \frac{1}{2}}{e}^{-r^2}]}_0^{\mathrm{\infty }}\\ \\ & =2\pi [-{\displaystyle \frac{1}{2}}{e}^{-(\mathrm{\infty }{)}^2}-(-{\displaystyle \frac{1}{2}}{e}^{-0^2})]\\ \\ & =2\pi [-{\displaystyle \frac{1}{2}}(0)-(-{\displaystyle \frac{1}{2}}(1))]\\ \\ & =2\pi {\displaystyle \frac{1}{2}}\\ \\ & =\pi \end{array}$‍

Isn't that a beautiful answer? It gets better, you can use this multi-dimensional result to solve the original single integral. Can you see how?