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Area of a Parabola


Date: 09/05/97 at 00:52:32
From: GLOBETROTTERS
Subject: Parabola

How do you find the area of a parabola? (I just finished Algebra 2.)

  c
  /\    <- The parabola looks something like this.
b{  }d
  \/       
  a

Point a is at (0,2) Point d is at (3,11) and the distance between 
point b and point d is 6.

What is the area of the parabola or, if you prefer, how can I find the
area of it?

I would much appreciate any response.


Date: 09/11/97 at 11:44:19
From: Doctor Rob
Subject: Re: Parabola

The picture you have drawn looks more like an ellipse. A parabola will
not be closed at the top, where you have shown point c. It would look
sort of like this:

\              /
 \            /
  `.        ,'
    `.    ,'
      `--'

Now if you draw a line bd from one side to the other, and a tangent 
line ef at the bottom tip a, and drop perpendiculars from b and d to 
ef, you will get a rectangle bdfe:

b________________d
|\              /|
| \            / |
|  `.        ,'  |
|    `.    ,'    |
|------`--'------|
e       a        f

The area inside the parabola bad and below the line bd is 2/3 of the 
area of the enclosing rectangle bdfe.

If the actual figure you meant to discuss is an ellipse, then the area
is Pi*r*s, where r and s are the semi-major and semi-minor axes' 
lengths (half of distance ac, apparently 9, and half of distance bd, 
apparently 3, so 27*Pi).

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 09/19/97 at 00:47:05
From: GLOBETROTTERS
Subject: Re: Parabola

Thank you. This is a great method, but is there a way of proving it?
(Can you prove that the area of "bad" is 2/3 of the area of rectangle
"bdef" ?)

Thank you in advance for a response.

Beginning Pre-Calculus student


Date: 09/19/97 at 10:16:28
From: Doctor Rob
Subject: Re: Parabola

There is if you understand limits or integral calculus. Below is the
proof using limits. The measurement of areas bounded by curves is not 
an easy exercise.  It is one of the reasons that calculus was 
developed.

Let the x-axis be eaf. Let the origin be a. Let the coordinates of 
point d be (A,B). Then the coordinates of b are (-A,B), those of e 
are (-A,0), and those of f are (A,0). Then the equation of the 
parabola has the form y = (B/A^2)*x^2.  

We observe that the parabola is symmetric about the y-axis, so we need 
only concern ourselves with the part of the region in the first 
quadrant, that is, to the right of point a, those with x- and
y-coordinates nonnegative. If that area is 2/3 the area of that part 
of the rectangle, then multiplying everything by 2 will establish the
relation we want. The area of that part of the rectangle is easily
seen to be A*B.

We will draw n-1 vertical lines between a and df, which are equally
spaced and distance apart of any two adjacent ones given by A/n.  
Their equations are x = k*A/n, where 0 < k < n.  Throw in the lines 
at a (x = 0, k = 0) and df (x = A, k = n), and you have cut the figure 
into n vertical slices of equal width. 

Now within each slice, there will be a part which is above the 
parabola and a part which is below the parabola. Take a typical 
slice, between x = (k-1)*A/n and x = k*A/n. If we draw a horizontal 
line segment between the vertical sides whose equation is
y = (B/A^2)*(k*A/n)^2 = B*k^2/n^2, then everything above it 
will form a smaller rectangle which lies within the region abd.  
If we add the areas of those smaller rectangles for each 
k = 1, 2, ..., n, we will get an area which is strictly smaller than 
the area we seek. The area of each smaller rectangle is its width A/n 
times its height B - B*k^2/n^2.  We have to add up

             n
   S(n) = Sum   [A/n*(B - B*k^2/n^2)]
             k=1

        = A*B - (A*B/n^3)*Sum k^2

        = A*B - (A*B/n^3)*n*(n+1)*(2*n+1)/6

        = A*B*[1 - (1+1/n)(2+1/n)/6]

        = A*B*[2/3 - 1/(2*n) - 1/(6*n^2)]

On the other hand, if we draw a horizontal line segment between the
vertical sides whose equation is y = B*(k-1)^2/n^2, then everything 
above it will form a smaller rectangle which includes all of the area 
of the region abd between the vertical sides. If we add the areas of 
those smaller rectangles for each k = 1, 2, ..., n, we will get an 
area which is strictly larger than the area we seek. The area of each 
smaller rectangle is its width A/n times its height B - B*(k-1)^2/n^2.  
We have to add up

             n
   T(n) = Sum   [A/n*(B - B*(k-1)^2/n^2)]
             k=1

        = A*B - (A*B/n^3)*Sum (k-1)^2

        = A*B - (A*B/n^3)*(n-1)*n*(2*n-1)/6

        = A*B*[1 - (1-1/n)(2-1/n)/6]

        = A*B*[2/3 + 1/(2*n) - 1/(6*n^2)]

So the actual area X satisfies

   S(n) < X < T(n).

Now as we let n -> infinity, you see that 

   lim S(n) = 2*A*B/3 = lim T(n)

since all the terms involving 1/n or 1/n^2 approach zero.  
Since X is caught between, it too must have value 2*A*B/3, 
which is 2/3 of the area of the rectangle (0,0)(A,0)(A,B)(0,B) 
enclosing it.  This is what we were trying to show.

The above argument is the long way of evaluating the definite 
integral from 0 to A of B - (B/A^2)*x^2 with respect to x, 
which is the proof using integral calculus.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Conic Sections/Circles
High School Geometry

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