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Area of an Ellipse without using Calculus


Date: 11/28/97 at 11:56:35
From: Vinay
Subject: How do you find the area of an oval

How do you find the area of an oval?


Date: 12/04/97 at 13:13:34
From: Doctor Mark
Subject: Re: How do you find the area of an oval

Hi Vinay,

I'm hoping that by "oval" you mean "ellipse," because that's the 
question I know the answer to.

If you're just interested in the answer, it's at the end of the 
message. If you want to know why, keep reading. Because yours was such 
a good question, it will take a little while to prove the result.

An ellipse is really just a circle that got stretched out in one 
direction, but not the other. Here's how you can think about it. Take 
a square sheet of rubber and draw a coordinate system on it, so that 
it looks like a rubbery sort of graph paper, all covered by squares, 
each square having a side of, say, 1 inch. Then draw a circle of 
radius R on this rubbery graph paper.

Now take opposite sides of this rubbery graph paper, and pull. I hope 
you can see that this would stretch both the little squares along one 
side, so that they look like rectangles, and the circle, so that it 
looks more oval than circular.

The oval shape is called an ellipse. If we take the "diameter" of the
ellipse in the direction of the stretch, that is called the "major 
axis" of the ellipse, and we'll let it be 2 times B, so that B is like 
the "radius" of the ellipse in the direction of the stretch. If we 
take the "diameter" of the ellipse along the direction perpendicular 
to the direction of stretch, that is called the "minor axis", and 
we'll let it be 2 times A, so that A is like the "radius" of the 
ellipse in the direction perpendicular to the stretch. In this case, 
since the circle didn't get stretched in that direction, A should be 
equal to R, the radius of the original circle. Remember that! We'll be 
using it shortly!

Now we could have calculated the area of the circle by multiplying the
number of squares (say N, where N would *not* be an integer) covered 
by the circle, times the area of each little square, which, since 
these little squares are 1 inch on a side, is 1 in^2. Note that N will 
not be an integer: the circle will not cover all the squares 
completely: some might only be covered up partially by the circle.

So the area of the circle could be given by

   Area of circle = N x (1 in^2) = N square inches

Now note that when we stretched the rubber sheet, the little squares 
on the graph paper just got stretched in one direction, just as the 
circle did. That is, each of the little squares got stretched into a 
rectangle. But notice also that the number of these rectangles inside 
the ellipse is the same as the number of squares inside the circle 
(this takes a little bit of thought).

We could find the area of the ellipse by multiplying this same number 
N by the are of each of the little rectangles. We know that the width 
of the rectangles didn't change (it's still just 1 inch), since the 
rubber sheet wasn't stretched in that direction. But the length is now 
a little bigger than an inch. How much bigger?

Suppose there were M little squares that were lying along a diameter 
of the original circle. All those M squares are now rectangles lying 
along the major axis of the ellipse. If we had added up the sides of 
those squares in the original circle, they would add up to the 
diameter (2R) of the original circle, so that

   M x (1 inch) = 2R

If we were to add up the lengths L of these M rectangles, they should 
add up to the major axis (2B) of the ellipse so that

   M x (L) = 2B

Now if we divide the second equation by the first, the M's cancel, and 
we find that

   L/(1 inch) = 2B/2R = B/R, i.e., L = (B/R) inches

Now remember, the area of each of these little rectangles is just its 
width (1 inch) times its length (B/R inches), i.e.,

   area of little rectangle = B/R square inches

There are N of these little rectangles in the ellipse, so the area of 
the ellipse should be

   area = N x (B/R) square inches

But remember that the area of the circle was N square inches. Hence,

   area of ellipse = 
   (B/R) x (N square inches) = 
   (B/R) x (area of the circle)

But the area of a circle is (pi)x(R^2), so we find that

   area of ellipse = (B/R)x(pi)x(R^2) = (pi)B(R^2/R) = (pi)BR

We're almost done...

Do you remember that the minor axis A of the ellipse was just equal 
to R, the radius of the original circle [because the circle didn't get 
stretched out in that direction]?  That means we can write:

   area of ellipse = (pi)AB

or, in words:

   the area of an ellipse is equal to pi times the product of the 
   major and minor axes.

Note that a circle is just an ellipse with its major and minor axes 
the same (both equal to R), so we get, for a circle:

   Area = (pi) RR = (pi)R^2

as we must.

I realize it may have taken some effort to follow this argument, but 
you should understand that one usually uses calculus to find the area 
of an ellipse, and so it takes a lot more work to get the answer 
without calculus. When you go on to take calculus, you will be able to 
do all of this in about three lines! (That's one reason calculus was 
invented: to make it easier to do things.)

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

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