Double integrals in polar coordinates

Background

• Polar coordinates (video)
• Double integrals beyond volume

What we're building to

• When you are performing a double integral,
  $\displaystyle \iint_\blueE{R} f\,\redE{dA}$\iint, start subscript, start color #0c7f99, R, end color #0c7f99, end subscript, f, start color #bc2612, d, A, end color #bc2612
  if you wish to express the function $f$f and the bounds for the region $\blueE{R}$start color #0c7f99, R, end color #0c7f99 in polar coordinates $(r, \theta)$left parenthesis, r, comma, theta, right parenthesis, the way to expand the tiny area $dA$d, A is

  $\redE{dA} = r\,d\theta\,dr$start color #bc2612, d, A, end color #bc2612, equals, r, d, theta, d, r

  (Pay attention to the fact that the variable $r$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 $R$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 $x^2 + y^2$x, squared, plus, y, squared.

Example 1: Tiny areas in polar coordinates

Warning!: You might be tempted to replace $\redE{dA}$start color #bc2612, d, A, end color #bc2612 with $d\theta\,dr$d, theta, d, r, since in cartesian coordinates we replace it with $dx\,dy$d, x, d, y. But this is not correct!

$\begin{aligned} \int_0^2 \int_0^{2\pi} r^3\,d\theta\,dr = \end{aligned}$

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[Answer]

Example 2: Integrating over a flower

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{aligned} \iint_R r\sin(\theta) \,dr\,d\theta \end{aligned}$

  A

  $\begin{aligned} \iint_R r\sin(\theta) \,dr\,d\theta \end{aligned}$
• (Choice B)  

  $\begin{aligned} \iint_R r^2 \sin(\theta) \,dr\,d\theta \end{aligned}$

  B

  $\begin{aligned} \iint_R r^2 \sin(\theta) \,dr\,d\theta \end{aligned}$

Check

[Answer]

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{aligned} \int_0^{2\pi} \int_0^{\cos(2\theta)} r^2 \sin(\theta) \,dr \,d\theta \end{aligned}$

  A

  $\begin{aligned} \int_0^{2\pi} \int_0^{\cos(2\theta)} r^2 \sin(\theta) \,dr \,d\theta \end{aligned}$
• (Choice B)  

  $\begin{aligned} \int_0^{\arccos(r)/2} \int_0^{2\pi} r^2 \sin(\theta) \,dr \,d\theta \end{aligned}$

  B

  $\begin{aligned} \int_0^{\arccos(r)/2} \int_0^{2\pi} r^2 \sin(\theta) \,dr \,d\theta \end{aligned}$

Check

[Answer]

$\begin{aligned} \int_0^{2\pi} \int_0^{\cos(2\theta)} r^2 \sin(\theta) \,dr \,d\theta = \end{aligned}$

Check

[Answer]

Example 3: The bell curve

Question: What is the integral $\begin{aligned} \int_{-\infty}^\infty e^{-x^2}\,dx \end{aligned}$ ?

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

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\begin{aligned} \int_{-\infty}^\infty \int_{-\infty}^{\infty} e^{-r^2}r\,d\theta\,dr \end{aligned}$

  A

  $\begin{aligned} \int_{-\infty}^\infty \int_{-\infty}^{\infty} e^{-r^2}r\,d\theta\,dr \end{aligned}$
• (Choice B)  

  $\begin{aligned} \int_0^\infty \int_0^{2\pi} e^{-r^2} r\,d\theta\,dr \end{aligned}$

  B

  $\begin{aligned} \int_0^\infty \int_0^{2\pi} e^{-r^2} r\,d\theta\,dr \end{aligned}$
• (Choice C)  

  $\begin{aligned} \int_0^\infty \int_0^{2\pi} e^{-r^2}\,d\theta\,dr \end{aligned}$

  C

  $\begin{aligned} \int_0^\infty \int_0^{2\pi} e^{-r^2}\,d\theta\,dr \end{aligned}$

Check

[Answer]

$\begin{aligned} \int e^{-r^2} r \, dr = \end{aligned}$

Check

[Answer]

$\begin{aligned} 2\pi \int_0^\infty e^{-r^2} r \, dr = \end{aligned}$

Check

[Answer]

Summary

• The only real thing to remember about double integral in polar coordinates is that

  $dA = r\,dr\,d\theta$d, A, equals, r, d, r, d, theta

  Beyond that, the tricky part is wrestling with bounds, and the nastiness of actually solving the integrals that you get. But those are the same difficulties one runs into with cartesian double integrals.
• The reason this is worth learning is that sometimes double integrals become simpler when you phrase them with polar coordinates, as was the case in the bell curve example.