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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.

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    $\begingroup$ Perhaps you should define $a$ and $b$. $\endgroup$ Mar 23, 2017 at 16:51
  • $\begingroup$ mathforum.org/library/drmath/view/54979.html $\endgroup$
    – user307169
    Mar 23, 2017 at 16:54
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    $\begingroup$ See proof 2: proofwiki.org/wiki/Area_of_Ellipse $\endgroup$
    – J126
    Mar 23, 2017 at 16:55
  • $\begingroup$ How are you going to define $\pi$? $\endgroup$
    – Jack M
    Mar 23, 2017 at 20:56
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    $\begingroup$ 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. $\endgroup$
    – Teepeemm
    Mar 23, 2017 at 21:39

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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 $\pi \cdot a \cdot b$.

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    $\begingroup$ @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. $\endgroup$
    – koe
    Mar 23, 2017 at 17:01
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    $\begingroup$ 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 $\endgroup$ Mar 23, 2017 at 17:14
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    $\begingroup$ 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. $\endgroup$ Mar 23, 2017 at 17:23
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    $\begingroup$ 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: 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. $\endgroup$ Mar 23, 2017 at 17:45
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    $\begingroup$ @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. $\endgroup$ Mar 24, 2017 at 0:33
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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 $\frac{\text{Area}(\text{circle})}{\left|\det\varphi\right|}=\pi a b$.

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    $\begingroup$ I like this idea! But how do you know that affine maps preserve the ratios between areas of non-polygons? $\endgroup$
    – wchargin
    Mar 23, 2017 at 23:31
  • $\begingroup$ @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. $\endgroup$ Mar 23, 2017 at 23:43
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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 $ab\pi$.

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  • $\begingroup$ I lost you , on why the resulting area is what it is. @Michael $\endgroup$
    – koe
    Mar 23, 2017 at 17:06
  • $\begingroup$ @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 $2a$ $\endgroup$
    – koe
    Mar 23, 2017 at 17:14
  • $\begingroup$ The area formula is a standard topic in linear algebra (it's the stretching argument that another answer made, but more formal). $\endgroup$ Mar 23, 2017 at 19:29
  • $\begingroup$ 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. $\endgroup$ Mar 23, 2017 at 19:39
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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.

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