# Area of an ellipse

1. May 12, 2015

### bigplanet401

1. The problem statement, all variables and given/known data
Use the parametric equations of an ellipse, x = f(t)= a cos t and y = g(t) = b sin t, 0 <= t <= 2 pi, to find the area that it encloses.

2. Relevant equations
Integral for parametric equations.

3. The attempt at a solution

$$A = \int_0^{2 \pi} g(t) f^\prime(t) \; dt = \int_0^{2 \pi} -ab sin^2 t \; dt = \frac{-ab}{2} \int_0^{2 \pi} (1 - cos 2t) \; dt = \frac{-ab}{2} (t - \frac{1}{2} sin 2t ) \vert_0^{2 \pi} = -\pi a b$$

But why the negative sign??

2. May 12, 2015

### Staff: Mentor

I think the reason is because you are essentially integrating in the reverse direction. To make things simple, let's look at just the area in the first quadrant, using thin vertical slices. That area (a quarter of the ellipse) is given by this integral : $\int_0^a y dx$. Here x will range from 0 to a.
When you replaced y and dx (and change limits of integration) you got $\int_0^{\pi/2} -ab sin^2 t \; dt$ (adjusting what you wrote to get only a quarter of the area).
As t ranges between 0 and $\pi/2$, the point on the ellipse moves from (a, 0) to (0, b), so x is moving from right to left along the x-axis (i.e., from x = a to x = 0), which will give you the opposite in sign as compared to when x moves from left to right.

3. May 12, 2015

### Zondrina

The parametric equations can be used in combination with the reverse implication of Green's theorem:

$$\frac{1}{2} \oint_C x dy - y dx$$

Find $dx$ and $dy$, then apply the parametrization.