Calculate area of Ellipse without calculus?

Question

I like the way integration works, but the final formula $\pi ab$ is too simple. I know there is a more deeper way to derive it. I just don't like to use calculus here, too many equations.

I'd like to use simple math, which does offer deeper insight into it.

In the answers, you keep coming back to an ellipse being the points where the sum of the distances from the foci is $2a$. Another way to define an ellipse is by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (which is equivalent to stretching a circle). Another way is by slicing a cone. Simple math without calculus means that you'll get various views of an ellipse, and that you are content with that. If you want an ellipse to be defined by the distance to its foci, it's not hard to show that gives the standard equation. But that's many equations. — Teepeemm, Mar 23 '17 at 21:39
Just an intuitive way to look at this formula: Area of ellipse is $\pi$ times the area of the circumscribing rectangle. For a circle, the rectangle becomes square. But again, why is this vague intuition valid relies on calculus. — FreezingFire, Mar 24 '17 at 05:00
As an aside, I don't know how to even define the notion of "area of an ellipse" without calculus. (or, at least, real analysis) — , Mar 24 '17 at 06:15
Wow xD "the final formula is too simple" ... said no mathematician EVER. Like, seriously, you sound disappointed xD — minseong, Mar 24 '17 at 07:57
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score 31 · Answer 1 · answered Mar 23 '17 at 16:55

31

Think about it this way. You start off with a circle of radius $a$ of which you know that it has area $\pi \cdot a^2$. Now you pick a direction (say horizontally for concreteness) and stretch the circle in that direction so that what used to be the diameter of length $2a$ will afterwards have length $2b$. Consequently, every line that lies horizontally will have been stretched by a factor of $b/a$, while you leave the vertical direction invariant. Then your total area will also have been changed by a factor of $b/a$, hence yielding $\pi \cdot a \cdot b$.

answered Mar 23 '17 at 16:55

Sebastian Schulz

988

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@Sebastian Then I lose the formal definition of ellipse, which is that the sum of the distances from the foci is $2a$ . How do I retain the definition of the ellipse. Also your proof has loop holes like that. So it's partial. – koe Mar 23 '17 at 17:01
@koe No, it's not partial. you refer to "a" definition of the ellipse. There are others. Sebastian Schulz is refering to another (equivalent) definition which amounts to say that an ellipse has equation $x^2/a^2+y^2/b^2=1.$ – Jean Marie Mar 23 '17 at 17:10
3

Well, I am not sure what loopholes you refer to. It was my understanding that you were asking for a more intuitive approach that circumvents more calculus.
As for your remark concerning the definition, that is an easy task in two-dimensional Euclidean geometry: Pick coordinates that put the foci at $(\pm c,0)$ and let your ellipse be the set of all points $(x,y)$ such that $\sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} = r$. Then you can convince yourself that $(\pm r/2, 0)$ satisfy that, so define $a=r/2$. Next, to find $b$ you'll need to find the points $(0, \pm b)$ that lie on the ellipse
– Sebastian Schulz Mar 23 '17 at 17:14
My post got too long, so here you go: You'll get an expression for $b$ in terms of $c$ and $r$. Then it will be easy to see that for any angle $\theta$, the point $(x,y) = (a \cos (\theta), b \sin (\theta) )$ is a solution, hence leading to the definition that I had assumed. For lack of space, I won't put down that calculation. – Sebastian Schulz Mar 23 '17 at 17:17
9

The problem with this is: how do you know (without calculus) that stretching the lines horizontally will change the area by the same factor? In fact, this is a very calculus-like statement, considering that the area consists of summing an infinitude of lengths. Cavalieri based his principle on a similar idea, but Archimedes eschewed it for a different proof based on more grounded concepts. – Paul Sinclair Mar 23 '17 at 17:23
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@PaulSinclair: You are absolutely right and I did in no way mean to suggest that my idea was a proof of any kind. I merely tried to suggest that one could think about it in a very visual way, though I certainly should have made that distinction clearer. Thank you for pointing out that sloppiness! – Sebastian Schulz Mar 23 '17 at 17:38
@Schulz you said such that \sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} = r – koe Mar 23 '17 at 17:39
but that is what we have to derive not prove. because we don't know the definition of the ellipse yet. And I am not interested in getting it's definition from the equation. Just like the equation of a circle is got from it's definition not vice-versa. – koe Mar 23 '17 at 17:40
I'd like to read the archimedes derivation too for area of an ellipse, where did you read that @Paul. – koe Mar 23 '17 at 17:42
3

I am not sure anymore what you are asking for. But you can find a lot of explanations for why the two definitions are indeed equivalent. See for example here: http://nebula.deanza.edu/~bloom/math43/ellipse-derivation.pdf which provides the implication in one direction (the one that I showed above). The implication in the other direction is much easier, you just have to plug it in. – Sebastian Schulz Mar 23 '17 at 17:45
I see what @PaulSinclair is saying: there is an aura of calculus if we're considering summing up the line segments. So, here we lack a derivation. Yet, as far as intuition goes, this is so good. Anyone familiar with the scale factors of similar figures should grasp the idea here. Actually, this begs me to ask, is there no such idea from Euclidean geometry - that scaling $n<N$ dimensions of any $N$ dimensional volume by a factor $s$ will scale the volume by $s^n$? This feels valid. Do we need linear algebra or calculus to show this? – zahbaz Mar 23 '17 at 22:22
@koe - I was actually referring to Archimedes proof for the volume of a sphere. He (apparently) first derived the fomula by means of a method that compared the areas of slices of the circle and a cone to that of a cylinder. But he did not consider that to be rigorous, so the proof he "published" was based on divide the sphere into many conical shapes with small bases (in its way also a precursor of calculus, but of the more rigorous type). I am not faulting Sebastion's post, by the way - I upvoted it - but I thought it best to point out this assumption. – Paul Sinclair Mar 23 '17 at 23:00
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@PaulSinclair, this is certainly not a rigorous presentation. Now, if we want to push in your direction, I'm forced to ask: "what is the definion (without calculus) of the area of a surface with curved edges?". I suspect (but don't know for sure) that that's a lot harder to answer rigorously. And I also suspect that any (calculus or not) working definition of area will trivially provide the behavior of the area under a dilation, as a corollary. – Euro Micelli Mar 24 '17 at 00:33
@EuroMicelli: It is indeed non-trivial. See this post for a brief explanation of some subtleties. – user21820 Mar 24 '17 at 02:36

score 11 · Answer 2 · answered Mar 23 '17 at 16:57

11

You may use an affine map $\varphi$ to send an ellipse into a circle. Since affine maps preserve the ratios between areas, the area of the ellipse is $\frac{\text{Area}(\text{circle})}{\left|\det\varphi\right|}=\pi a b$.

answered Mar 23 '17 at 16:57

Jack D'Aurizio

353,855

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I like this idea! But how do you know that affine maps preserve the ratios between areas of non-polygons? – wchargin Mar 23 '17 at 23:31
@wchargin: the key idea behind the definition of area through the Peano-Jordan measure is that any shape deserving an area can be decomposed into an almost-disjoint union of polygons. The main technicality is that such union is not always the union of a finite amount of pieces, but the outlined principle applies to every piece in the same way. – Jack D'Aurizio Mar 23 '17 at 23:43

score 4 · Answer 3 · answered Mar 23 '17 at 16:56

4

Consider the unit disk (bounded by the circle of radius $1$, centered at the origin). Now, to construct an ellipse whose axes are $a$ along the $x$-axis and $b$-along the $y$-axis. This corresponds to the application of the linear transformation $$ \begin{bmatrix}a&0\\0&b\end{bmatrix}. $$ We can confirm that this is an ellipse because if your original coordinates are $x_1$ and $x_2$ while your new coordinates are $y_1$ and $y_2$, we have $y_1=ax_1$ and $y_2=bx_2$. Therefore, $y_1$ and $y_2$ satisfy: $$ \frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}=1. $$

Since linear transformations scale areas by the determinant (and the original disk has area $\pi$), the resulting area is $ab\pi$.

answered Mar 23 '17 at 16:56

Michael Burr

32,867

I lost you , on why the resulting area is what it is. @Michael – koe Mar 23 '17 at 17:06
@Burr but how do we know easily that it is an ellipse. Not the equation of an ellipse, but that the sum of the distance from the points to the two foci is $2a$ – koe Mar 23 '17 at 17:14
The area formula is a standard topic in linear algebra (it's the stretching argument that another answer made, but more formal). – Michael Burr Mar 23 '17 at 19:29
For the way to see that this is an ellipse, you might need to review how to connect the formula for an ellipse with its geometric meaning. – Michael Burr Mar 23 '17 at 19:39

score 0 · Answer 4 · answered Mar 24 '17 at 06:05

Draw a rectangle of known area around your shape. Start throwing a known number of darts at the rectangle randomly, the more the better. Count the number of darts landed inside the boundaries of your shape versus those outside, but still within the boundaries of the rectangle. Apply that ratio to the known area of the rectangle, and there you have an approximation of the size of your shape.

Calculate area of Ellipse without calculus?

4 Answers4