You are about to erase your work on this activity. Are you sure you want to do this?

Updated Version Available

There is an updated version of this activity. If you update to the most recent version of this activity, then your current progress on this activity will be erased. Regardless, your record of completion will remain. How would you like to proceed?

Mathematical Expression Editor

Chat

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.

Area of a Polar Region Let be continuous and non-negative on , where . The area of
the region bounded by the curve and the lines and is

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

Compute the area of one petal of the polar curve :

From the picture it looks like integrating from to will give us the area of our desired
region. We can convince ourselves that this is correct by inspecting and noting that
our curve starts at in the -plane when , and then moves to the origin. Since the first
positive value of that make is , we see that our eyes did not lie to us. Write with me

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 mistakea complete curve is
drawn when runs from to

A good way to understand the polar graph is to graph along side of :

Areas between polar curves

Consider the shaded region:

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

Area Between Polar Curves The area of the region bounded by and , and , where
on , is

Find the area of the region that lies within but not within

This corresponds to the
following region:

Hence to find the are we compute

When computing regions between polar curves, one must be careful. Sometimes, one
must compute the points of intersection of the two curves, and this can be difficult.

Find
the points of intersection between the curves

To find the points of intersection of two polar curves, the best thing to do is to look
at a graph.

From the graph above, we see that there are points of intersection. Let’s try to use
algebra to find them. Write with me

Ah, and this is true when: However, from the graph we see there is one more point of
intersection. Namely When you are unable to look at a graph, you can
still use algebra to try and find the points of intersection, but some may be
hidden.

Find the area of the region that lies within both polar curves

To compute this area, we’ll use the previous example. First we’ll compute the area of
this region:

So to compute the region we can either compute: Evaluating either integral and
multiplying by tells us the area is .