See Theorem 14.3, pg.1003[ Note: The continuity information in the theorem will be present in the problems we solve and can be taken to mean no pathologies are present that might further complicate the evaluations.]

Double ( later in triple integral problems ) Integrals are often much easier to compute if we use a polar coordinate representation for the function and the region under discussion. Theoren 14.3, pg.1003 contains the information we need to make the variable changes. Now we will review some of the previous information you've learned concerning polar coordinates and apply them, along with Thm. 13.3, to compute some double integrals.

See 10.4 Thm 10.10, pg. 730 Coordinate Conversion

If we locate points in a region by their polar coordinates then in the Iterated Integral we need a formula in polar coordinates for the function under discussion. To get such a formula we will substitute for x and y from the conversion equations.

We also need to consider what to do with the Iterated Integral now that we're using polar coordinates to describe the region. It should be noted that the dA notation used in the Double Integral symbol makes more sense after we see the reasoning behind using in the iterated integral.
A simple explanation ( perhaps too simple ) would be to remind you that when the integrand of a Double Integral is constant-valued at f(x,y)=1 the value of the integral relates to area measure of a region defined by the limits of integration and that area measure would also be the measure of volume of a solid with constant height of one unit. [We used this idea in previous notes on Double versus Iterated Integrals.]

If you recall that the measure of area for a sector of a circle of radius R and central angle measure 

(in radians) is , then the iterated integral to measure that area should read [where the difference of theta and beta is the angle measure in radians]. This yields which is consistent with the afore mentioned formula for area measure of a sector.

A more detailed explanation is as follows:

See Exercises 14.3 Problem 61

See Exercises 14.3 Problem 61 solution

The explanation for the formula for the measure of area of the sector of a circle can be done by reasoning that the area of the sector is to the area of the circle of radius R as the measue of the central angle of the sector is to 2. Try writing that idea to get the formula .

Now we can solve a problem using Theorem 13.3

See 14.3 Example 3

See 1.3 Example 3 Solution

Look at the following solved problems in the Exercises at the end of 14.3 : 1,3,13,15,21

Then do the following unsolved problems as homework: 2,4,16,18,22,24