Polar Coordinates + Double Integrals

Takeaway: 

For a polar region R bounded by two polar curves r1(θ) and r2(θ) and between θ1=α and θ2=β, we have

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

$f(r, \theta) = r^2$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

$r \le 2$r, is less than or equal to, 2

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

It's not $d\theta$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 $r$r.

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

$\redE{dA} = (r\,d\theta)(dr)$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

$\begin{aligned} \iint_{r\le 2} r^2\,\redE{dA} = \iint_{r\le 2} r^2\,(r\,d\theta)(dr) = \iint_{r\le 2} r^3\,d\theta\,dr \end{aligned}$

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