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Double integrals in polar coordinates

If you have a two-variable function described using polar coordinates, how do you compute its double integral?

What we're building to

  • When you are performing a double integral,
    \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 and the bounds for the region start color #0c7f99, R, end color #0c7f99 in polar coordinates left parenthesis, r, comma, theta, right parenthesis, the way to expand the tiny area d, A is
    start color #bc2612, d, A, end color #bc2612, equals, r, d, theta, d, r
    (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, squared, plus, y, squared.

Example 1: Tiny areas in polar coordinates

Suppose we have a multivariable function defined using the polar coordinates r and theta,
f, left parenthesis, r, comma, theta, right parenthesis, equals, r, squared
And let's say you want to find the double integral of this function in the region where
r, is less than or equal to, 2
This is a disc of radius 2 centered at the origin.
Written abstractly, here's what this double integral might look like:
r2r2dA\begin{aligned} \iint_{r\le 2} r^2\,\redE{dA} \end{aligned}
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 start color #bc2612, d, A, end color #bc2612 term?
Warning!: You might be tempted to replace start color #bc2612, d, A, end color #bc2612 with d, theta, d, r, since in cartesian coordinates we replace it with d, x, d, y. 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 start color #bc2612, d, A, end color #bc2612 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 d, r, a tiny change in the r-coordinate.
Using a differential d, r 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 d, r and d, theta each approach 0 it basically is a rectangle, its area is
start color #bc2612, d, A, end color #bc2612, equals, left parenthesis, r, d, theta, right parenthesis, left parenthesis, d, r, right parenthesis
Plugging this into our original integral, we get
r2r2dA=r2r2(rdθ)(dr)=r2r3dθdr\begin{aligned} \iint_{r\le 2} r^2\,\redE{dA} = \iint_{r\le 2} r^2\,(r\,d\theta)(dr) = \iint_{r\le 2} r^3\,d\theta\,dr \end{aligned}
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 d, r, 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
0202πr3dθdr=\begin{aligned} \int_0^2 \int_0^{2\pi} r^3\,d\theta\,dr = \end{aligned}

Example 2: Integrating over a flower

Define a two-variable function f in polar coordinates as
f, left parenthesis, r, comma, theta, right parenthesis, equals, r, sine, left parenthesis, theta, right parenthesis
Let R be flower-shaped region, defined by
r, is less than or equal to, cosine, left parenthesis, 2, theta, right parenthesis
Solve the double integral
RfdA\begin{aligned} \iint_R f\,dA \end{aligned}
Step 1: Which of the following represents the right way to replace f, d, A in the abstractly written double integral?
Choose 1 answer:

Step 2: Now we must encode the fact that R is defined as the region where r, is less than or equal to, cosine, left parenthesis, 2, theta, right parenthesis. Which of the following is the right way to put bounds on the double integral?
Choose 1 answer:

Step 3: Solve this integral.
02π0cos(2θ)r2sin(θ)drdθ=\begin{aligned} \int_0^{2\pi} \int_0^{\cos(2\theta)} r^2 \sin(\theta) \,dr \,d\theta = \end{aligned}

Example 3: The bell curve

Are you ready for one of my favorite results in math? This is really quite clever.
Question: What is the integral ex2dx\begin{aligned} \int_{-\infty}^\infty e^{-x^2}\,dx \end{aligned} ?
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 x, y-plane.
xy-planee(x2+y2)dA\begin{aligned} \iint_{xy\text{-plane}} e^{-(x^2 + y^2)} \,dA \end{aligned}
If we keep everything in cartesian coordinates, this is as hard to solve as the original single integral.
e(x2+y2)dxdy\begin{aligned} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} \,dx\,dy \end{aligned}
However, something magical happens when we convert to polar coordinates.
Concept check: Express this double integral using polar coordinates.
Choose 1 answer:

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:
002πer2rdθdr=0er2r02πdθThis evaluates to 2πdr=0(er2r)(2π)dr=2π0er2rdr\begin{aligned} &\quad \int_0^\infty \int_0^{2\pi} e^{-r^2} r \,d\theta \,dr \\\\ &= \int_0^\infty e^{-r^2} r \underbrace{ \int_0^{2\pi} \,d\theta }_{\text{This evaluates to $2\pi$}}\,dr \\\\ &= \int_0^\infty \left(e^{-r^2} r \right)(2\pi) \, dr \\\\ &= 2\pi \int_0^\infty e^{-r^2} r \, dr \end{aligned}
Concept check: Find the antiderivative of e, start superscript, minus, r, squared, end superscript, r, using either u-substitution or the inverse chain rule.
er2rdr=\begin{aligned} \int e^{-r^2} r \, dr = \end{aligned}

Notice, the reason you can now find an antiderivative is because of that little r term that showed up due to the fact that d, A, equals, r, d, theta, d, r.
Concept check: Using this antiderivative, finish solving the integral which computes volume under a three-dimensional bell curve.
2π0er2rdr=\begin{aligned} 2\pi \int_0^\infty e^{-r^2} r \, dr = \end{aligned}

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?


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

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