Area of an Ellipse without using CalculusDate: 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/ |
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