Why is the area of an ellipse negative?

[bigplanet401]

Homework Statement

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.

Homework Equations

Integral for parametric equations.

The Attempt at a Solution

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

But why the negative sign??

 

[Mark44]

Homework Statement

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.

Homework Equations

Integral for parametric equations.

The Attempt at a Solution

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

But why the negative sign??

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.

 

[STEMucator]

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.