13. Tickle your student’s fancy – and answer the question, “What’s all this Calculus good for, anyway?”
As a tutor, if you memorize the solution method to this problem and show it to your disenfranchised Calculus student, it may help them to see how this material can be applied in everyday life.
Where did the formula
(area of a circle) come from?
The general equation of a circle (of radius r, centered at
the origin) is ![]()
Solve for y (to express it as a function):
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This describes the part of a circle that is in the first quadrant. Its area is ¼ that of a full circle. See the diagram below.

If we can integrate the function and find the area underneath the curve between 0 and radius r, then we will be able to multiply the result by 4 to get the formula for the area within a full circle of radius r.
So the problem to solve is this:
![]()
This is best solved by trigonometric substitution.
Let ![]()
Let ![]()
Limits of integration become 0 to
which represents the 1st quadrant, expressed in
radian measure. To begin, substitute for
and
, and change the limits of integration:
![]()
![]()
![]()
![]()
![]()
![]()
![]()
Now use a half-angle formula:
![]()
![]()
Separate the fractions.






Hey, I remember this!