Recall the relation between the rectangular coordinates and polar coordinates is

Now consider the part of grid obtained by fixing θ at various values and varying r and then by fixing r at various values and varying θ.
The idea is that these lines obtained by fixing one or the other coordinate are very close together, much closer than drawn and so we would expect the area of one of the little curvy quadrilaterals to be close to the area of the parallelogram shown. Consider this parallelogram. The two sides originating at the intersection of two of the grid lines as shown are approximately equal to

where dr and dθ are the respective small changes in the variables r and θ. Thus the area of one of those little curvy shapes should be approximately equal to

by the geometric description of the cross product. These vectors are extended as 0 in the third component in order to take the cross product. This reduces to

which is the increment of area in polar coordinates, taking the place of dxdy. The integral is really about taking the value of the function integrated multiplied by dA and adding these products. Here is an example.
Example 15.1.1 Find the area of a circle of radius a.
The variable r goes from 0 to a and the angle θ goes from 0 to 2π. Therefore, the area is

Example 15.1.2 The density equals r. Find the total mass of a disk of radius a.
This is easy to do in polar coordinates. The disk involved has θ going from 0 to 2π and r from 0 to 2. Therefore, the integral to work is just

Notice how in these examples the circular disk is really a rectangle
Example 15.1.3 Find the area of the inside of the cardioid r = 1 + cosθ, θ ∈
Here the integral is

To see how impossible this problem is in rectangular coordinates, draw the graph of the cardioid.
How would you go about setting this up in rectangular coordinates? It would be very hard if not impossible, but is easy in polar coordinates. This is because in polar coordinates the region integrated over is the region below the curve in the following picture.
Example 15.1.4 Let R denote the inside of the cardioid r = 1 + cosθ for θ ∈

Here the convenient increment of area is rdrdθ and so the integral is

Now you need to change x to the right coordinates. Thus the integral equals

A case where this sort of problem occurs is when you find the mass of a plate given the density.
Definition 15.1.5 Suppose a material occupies a region of the plane R. The density λ is a nonnegative function of position with the property that if B ⊆ R, then the mass of B is given by ∫ _{B}λdA. In particular, this is true of B = R.
Example 15.1.6 Let R denote the inside of the polar curve r = 2 + sinθ. Let λ = 3 + x. Find the total mass of R.
As above, this is
