Problem:
Find the area of the region which lies inside the graph of

and outside of the graph of
.
Visualization:
Using JKGraph:
- Click on the XY button so that it changes to Polar.
- On the Formulas menu, choose Polar Formula: R=F(@).
- At the prompt R(@) = , type in
1 + cos(@)
Note that the symbol @ is used to denote the angle.
- Move to the prompt @ Starting Value and enter -180.
In this program the angle is entered in degrees.
- Move to the prompt @ Ending Value and enter 180.
- Click on OK and then you will see the graph of the cardiod.
- Click on the 2 Off button so that we can enter two functions.
The default second function is then graphed.
- We want to change the default second function and so we click on
the 1 - 2 button before we can make this change.
- Again, on the Formulas menu, choose Polar Formula: R=F(@).
- At the prompt R(@) = , type in
3*cos(@)
- Move to the prompt @ Starting Value and enter -180.
- Move to the prompt @ Ending Value and enter 180.
- Click on OK and then you will see the graph of the circle together
with the cardiod which we entered above.
- We calculate that the points of intersection of the two graphs; this
can be done using this program (see ??). By whatever technique we use, we
have that the intersections occur at the angles of -60 and 60 degrees and
at R = 0.
- On the Domains menu, click on Integral Domain.
- Move to the prompt Integral Upper Limit: and enter -60.
- Move to the prompt Integral Lower Limit: and enter 60.
- Move to the prompt Integral Method and choose one of the first
three: lower, midpoint or upper Riemann sums.
- Click on OK.
- Now click on the integral button and you will see the approximating
wedges. The numerical value of the approximate area appears on the
bottom bar.