
We integrate polar functions.

When using rectangular coordinates, the equations $$ and $$ defined vertical and horizontal lines, respectively, and combinations of these lines create rectangles (hence the name ‘‘rectangular coordinates’’). It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral.

When using polar coordinates, the equations $$ and $$ form lines through the origin and circles centered at the origin, respectively, and combinations of these curves form sectors of circles. It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles. Recall that the area of a sector of a circle with radius $$ subtended by an angle $$

is $$. So given a polar plot, partition the interval $$ into $$ equally spaced subintervals as $$:
The length of each subinterval is $$, representing a small change in angle. The area of the region defined by the $$th subinterval $$ can be approximated with a sector of a circle with radius $$, for some $$ in $$. The area of this sector is $$. This is shown here
where $$ has been divided into $$ subintervals. We approximate the area of the whole region by summing the areas of all sectors: This is a Riemann sum! By taking the limit of the sum as $$, we find the exact area of the region in the form of a definite integral.

The theorem states that $$. This ensures that region does not overlap itself, giving a result that does not correspond directly to the area.

Now I present you with a mystery: You know that the area of one petal of $$ has area $$. Hence, all three petals must have an area of $$. Compare this to

How is it possible that the total area of the three petals is $$, but the integral above is $$?

one (or more!) of our computations has a mistake $$ a complete curve is drawn when $$ runs from $$ to $$

### Areas between polar curves

If we let $$ represent the circle, and $$ represent the cardioid, we can find the area of this region by computing the area bounded by $$ and subtracting the area bounded by $$ on $$. Thus