Question 07 (a)
Show that the area of the region inside the ellipse
where a and b are positive constants, equals
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.
First isolate for y.
Find the points of intersection of the ellipse and the x-axis.
Note that by symmetry we need to only consider the positive square root and double the resulting value.
Either interpret this result as an area or try the trig substitution
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
We proceed as in the hints. First, we isolate for y to see that
Taking the square roots yields
Now, we want to know the aera fo the ellipse. First, we set to find the points of intersection with the x-axis. This gives
So . Thus, we wish to integrate
from to . By symmetry, we need only to integrate the positive function above and then double our resulting value (pictorially, the ellipse has a symmetry that is reflection about the x-axis. So to get the full area, compute half of it then double your result). Thus, we have that
This last integral can be interpreted as the area of a half circle of radius . This has area . Thus, our area is
completing the question.
We proceed as in solution one until the line
Now we do a trig substitution. Let so that . Changing the bounds, we see that so and so . Then the above becomes
Now we use the double angle formula
to see that