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

and outside of the graph of

.

Visualization:
Using JKGraph:

  1. Click on the XY button so that it changes to Polar.
  2. On the Formulas menu, choose Polar Formula: R=F(@).
  3. At the prompt R(@) = , type in
    1 + cos(@)
    Note that the symbol @ is used to denote the angle.
  4. Move to the prompt @ Starting Value and enter -180. In this program the angle is entered in degrees.
  5. Move to the prompt @ Ending Value and enter 180.
  6. Click on OK and then you will see the graph of the cardiod.
  7. Click on the 2 Off button so that we can enter two functions. The default second function is then graphed.
  8. We want to change the default second function and so we click on the 1 - 2 button before we can make this change.
  9. Again, on the Formulas menu, choose Polar Formula: R=F(@).
  10. At the prompt R(@) = , type in
    3*cos(@)
  11. Move to the prompt @ Starting Value and enter -180.
  12. Move to the prompt @ Ending Value and enter 180.
  13. Click on OK and then you will see the graph of the circle together with the cardiod which we entered above.
  14. 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.
  15. On the Domains menu, click on Integral Domain.
  16. Move to the prompt Integral Upper Limit: and enter -60.
  17. Move to the prompt Integral Lower Limit: and enter 60.
  18. Move to the prompt Integral Method and choose one of the first three: lower, midpoint or upper Riemann sums.
  19. Click on OK.
  20. 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.