# I Area ellipse: parametric form, angles and coincidences

1. Mar 9, 2017

### haushofer

Dear all,

I have a question regarding the computation of the area of an ellipse. The parametric form of the ellipse with axes a and b is

$$x(t) = a\cos{(t)}, \ \ \ y(t) = b\sin{(t)}$$

Using this to evaluate the area of the ellipse, usually one takes one halve or one quarter of the ellipse and uses symmetry. E.g.,

$$\frac{1}{4} \text{area} = \int_{x=0}^{x=a} y(x) dx = -ab \int_{t=\frac{\pi}{2}}^{t = 0} \sin^2{(t)} dt = \frac{\pi ab}{4}$$

My question is as follows: if one does the calculation as follows,

$$\text{area} = ab \int_{t=0}^{t = 2\pi} \sin^2{(t)} dt = \pi ab$$

one obtains the right answer. However, one integrates from one point on the ellips to the very same point, and the integrand is also not positive for half of the integral, so I can see that integration from t =0 to t = 2 pi is not allowed. Is it then a coincidence that one still obtains the right answer?

2. Mar 9, 2017

### Mastermind01

This is what I think:

As you parameterized the equation of the ellipse, while doing the integral you are finding out the area under the curve of the function $sin^2(t)$ which is always positive and for which the constraints (which are present for integrating a regular ellipse equation) are not present. Thus , you can integrate fully to get the correct answer because you are no longer integrating from the same point to the same point as the function under consideration has changed.

3. Mar 9, 2017

### vanhees71

To be on the safe side, you should really calculate an area integral by parametrizing the entire area. One convenient one is
$$\vec{x}(\lambda,\phi)=\lambda (a \cos \phi,b \sin \phi).$$
Then obviously the entire area is parametrized by $\lambda \in [0,1]$, $\phi \in [0,2 \pi)$. The area element is
$$\mathrm{d}^2 A=\mathrm{d} \lambda \mathrm{d} \phi \left |\mathrm{det} \frac{\partial(x,y)}{\partial(\lambda,\phi)} \right| = \mathrm{d} \lambda \mathrm{d} \phi \mathrm{det} \begin{pmatrix} a \cos \phi & b \sin \phi \\ -a \lambda \sin \phi & b \lambda \cos \phi \end{pmatrix} =ab\lambda,$$
and thus for the area you get
$$A=\int_0^{1} \mathrm{d} \lambda \int_0^{2 \pi} \mathrm{d} \phi ab \lambda=2 \pi ab \int_0^1 \mathrm{d} \lambda \lambda = \pi ab.$$

4. Mar 9, 2017

### haushofer

I agree, but I'm teaching a calculus course now where these kind of area parametrisations are not discussed. :P

5. Mar 9, 2017

### haushofer

I don't see this; the parametric form implies that

$$(x(t=0),y(t=0)) = (x(t=2\pi),y(t=2\pi))= (a,0)$$

So effectively you're integrating from (a,0) to (a,0). But I see that the integrand is a square, and hence is positive for $t \in [0,2\pi]$.

6. Mar 9, 2017

### haushofer

Or am I mistaken and is it allowed to calculate the area enclosed by a closed parametric curve like an ellipse by evaluating the integral from t=0 to t=2pi?

7. Mar 9, 2017

### vanhees71

I'm not sure I understand the arguments used to evaluate the area as "area under a curve" $y=f(x)$ at all. Here, to be safe you have to first write the curve as function of $x$. Of course, you can do this only for a part of the ellipse. Taking the upper part you have
$$f(x)=b \sqrt{1-x^2/a^2}.$$
Thus you get
$$A=2\int_{-a}^{a} \mathrm{d} x f(x),$$
where the factor of $2$ comes from taking both halves of the ellipse to get the full area.

To evaluate the integral you can of course do the standard substitution
$$x=a \cos t, \quad t \in [0, \pi], \quad \mathrm{d} x = -\mathrm{d} t a \sin t.$$
Then you get
$$A=2 ab \int_0^{\pi} \mathrm{d} t \sin^2 t=ab \int_{0}^{\pi} \mathrm{d} t [1-\cos(2 t)] = a b \left (t-\frac{1}{2} \sin 2 t \right)_{t=0}^{t=\pi}=\pi a b.$$

8. Mar 9, 2017

### Mastermind01

No, it's not allowed if you express y as function of x and then integrate because of sign constraints. However by changing it into a parametric equation, the integral became the area under a different curve $sin^2 x$ which unlike an ellipse "goes on." So, I think that's why the two areas turned out to be the same. In general though the area is done by @vanhees71 's method.

9. Mar 9, 2017

### PeroK

By integrating the positive function (above the x-axis) from $x=0$ to $x=a$ you have a positive $dx$ and a positive integrand. You could then integrate the negative function (below the x-axis) from $x = a$ to $x = -a$ and you would get a postive result, as in this case $dx$ is negative. Finally, you would integrate the other positive quarter from $x=-a$ to $x =0$ and again get a positive result. Hence:

$A = \int_{0}^{a} y_1(x) dx + \int_{a}^{-a} y_2(x)dx + \int_{-a}^{0} y_1(x) dx$

Where $y_1, y_2$ are positive and negative respectively. The middle integral is positive because you are integrating "right to left".

And, using the $t$ substitution in these integrals you get:

$A = \int_{\pi/2}^{0} (-ab \sin^2(t))dt + \int_{2\pi}^{\pi} (-ab \sin^2(t))dt + \int_{\pi}^{\pi/2} (-ab \sin^2(t))dt = \int_{0}^{2\pi} (ab \sin^2(t))dt$

So, there is no coincidence and it's perfectly valid.

10. Mar 11, 2017

### haushofer

Thanks! That makes sense. I must admit I'm a bit confused now as to how general this result is (i.e. does integration over a closed contour of a parametric eqn. always give you the area). In all the examples I can find, the integral is done partly and then symmetry is used.

11. Mar 11, 2017

### PeroK

In general, it doesn't depend on the particular functions $y_1$ (above the x-axis) and $y_2$ (below the x-axis), or on the orientation of the parameterisation - as it's just a regular integral substitution. Nor does it need to be a closed curve with a periodic parametrisation.

All that you really have here is a way to add two areas (one above the x-axis and one below the axis) by taking a negative $dx$ and leaving the second function negative.

The critical factor, therefore, is the behaviour of the curve as it crosses the x-axis. That determines how you use integration to get area in the first place.

Last edited: Mar 11, 2017