Double Integrals in Polar Coordinates

See Polar Coordinates, Integration of Scalar Fields

The Riemann Sum in Polar Coordinates:

(See Riemann Sum)

So just remember that is the radius and is the angle from the x-axis.

$$ \begin{align} \Delta A_{i} & = \frac{1}{2} \rho_{i}^2 \Delta\varphi - \frac{1}{2} \rho^2_{i - 1} \Delta \varphi \\ & = \frac{1}{2} (\rho_{i}^2 - \rho_{i - 1}^2) \Delta \varphi \\ & = \frac{1}{2} (\rho_{i} + \rho_{i - 1})(\rho_{i} - \rho_{i - 1}) \Delta\varphi && \text{Difference of squares} \\ & = \ubrace{\frac{1}{2} (\rho_{i} - \rho_{i - 1})}_{\text{Average}} \Delta \rho \Delta\varphi \\ dA & = \rho_{i}^* \Delta \rho\Delta\varphi \end{align} $$
Definition

In polar coordinates, the area of a function is:

You may sometimes see this as (mostly in E&M, where is reserved for charge density), or (mostly with math).

Equation

Example

Find the volume enclosed beneath and

this will be easier in Polar Coordinates.

Example

Find the area contained in one leaf of :

solution

Example

Evaluate , where is the portion of the unit circle in the first quadrant.

solution

Note that we have , which in polar coordinates, will be , so

Example

Sketch the domain of integration of the integral below and evaluate it by transforming to polar coordinates.

solution
Our coordinate goes from to and our coordinate goes from to . If we re-write this, we get:

I'm honestly not sure how you're supposed to come up with that last bit.

This equation describes the equation of a circle with radius centred at .

Now recall that and .

For integrating on , we go from the line to the circle . In polar form, . For the circle, we have:

From the diagram, we also have .

Putting it together (holy shit):

Now we solve by substitution. Let .