dA = r dr d theta

Conceptually, computing double integrals in polar coordinates is the same as in rectangular coordinates. After all, the idea of an integral doesn't depend on the coordinate system. If $R$ is a region in the plane and $f(x,y)$ is a function, then $\iint_R f(x,y) dA$ is what we get when we

Chop $R$ into a bunch of small pieces.
Compute $f(x^*,y^*)\Delta A$ for each piece, where $\Delta A$ is the area of the piece.
Add up the contributions of the pieces, and
Take a limit as we chop $R$ into smaller and smaller pieces.

The differences when working with polar coordinates are explained in the following video:

When working with rectangular coordinates, our pieces are boxes of width

$\Delta x$ , height

$\Delta y$ , and area

$\Delta A = \Delta x \Delta y$ . When working with polar coordinates, our pieces are polar rectangles. These are regions of the form

$\theta \in [\alpha, \beta]$ ,

$r \in [a,b].$ The approximate contribution of a single box to our integral is

$f(r_i^*,\theta_j^*) \Delta A$ , where

$\Delta A$ is the area of the box. [Here we are thinking of

$f$ as a function of

$r$ and

$\theta$ . If

$f$ is a function of

$x$ and

$y$ , then strictly speaking we should write

$f(r \cos(\theta), r\sin(\theta))$ instead of

$f(r,\theta)$ .]

So what is the area of a polar rectangle? The polar rectangle is the difference of two pie wedges, one with radius $b$ and one with radius $a$ , and so has area

$\begin{eqnarray*} \Delta A & = & \frac{b^2}{2}(\beta-\alpha) - \frac{a^2}{2}(\beta - \alpha) \\ & = & \frac{b^2-a^2}{2}(\beta-\alpha) \\ & = & \frac{a+b}{2} (b-a)(\beta - \alpha) \\ & = & r^* \Delta r \Delta \theta, \end{eqnarray*}$ where

$r^*=\frac{a+b}{2}$ is the average value of

$r$ ,

$\Delta r = b-a$ is the change in

$r$ and

$\Delta \theta = \beta-\alpha$ is the change in

$\theta$ .

Another way to understand this is to look at the shape of a small polar rectangle, like the shaded box in the figure above. This is almost a rectangular, only rotated and with slightly curved sides. One side has length $\Delta r$ . The other had length approximately $r \Delta \theta$ . So the whole rectangle has approximate area $r \Delta r \Delta \theta$ .

Now we organize our boxes. Since the contribution of each box is approximately

$f(r_i^*,\theta_j^*) r_i^* \Delta r \Delta \theta$ , the contribution of the boxes in the same thin pie wedge is approximately

$\left (\int_a^b f(r,\theta_j^*) r dr \right ) \Delta \theta$ .

Finally we put all the pie wedges together, to get a total of

$\int_{\alpha}^\beta \left ( \int_a^b f(r,\theta) r dr \right ) d\theta$ .

The infinitestimal area in rectangular coordinates is

$dA = dx\, dy$ . The infinitesimal area in polar coordinates is

$dA = r \, dr \, d\theta.$ Don't forget the factor of $r$ !!

Double integrals in polar coordinates become iterated integrals

$\iint f(x,y) dA \ = \ \int\int f(r\cos(\theta), r\sin(\theta)) r \, dr \, d\theta \ = \ \int \int f(r\cos(\theta), r\sin(\theta)) r\, d\theta\,dr,$ with appropriate limits of integration.

Integration by Parts

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Calculus with Parametrized Curves

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