Negative sign in calculating area of an ellipse using parametric equations

Question

I'm trying to prove the formula for the area of an ellipse using parametric equations. If we let $x(t)=a\ cos(\varphi)$ and $y(t)=b\ sin(\varphi)$, then $dx = -a\ sin(\varphi)\ d\varphi$. So:

$$\int y\ dx = -ab\int_{0}^{2\pi} \sin^2(\varphi)\ d\varphi$$

I evaluated the integral and I got $-ab\pi$. Why is there a negative sign? Am i doing something wrong? Do I need to allays take the absolute value if the area is negative or that just means that I'm doing it wrong?

Answer 1

Had you used the following symmetrical formula (see explanation below)

$$\int_{\varphi=0}^{\varphi=2\pi}\tfrac12(x dy - y dx),\tag{1}$$

you wouldn't have even noticed the problem. Indeed, one obtains in this way :

$$\int_{\varphi=0}^{\varphi=2\pi}\tfrac12 ab( \underbrace{(\cos \varphi)^2 + (\sin \varphi)^2)}_{= \ 1}d\varphi = \pi ab$$

Explanation of (1): the elementary area can be computed as :

$$\tfrac12\begin{vmatrix}x&x+dx\\y&y+dy\end{vmatrix}=\tfrac12\begin{vmatrix}x&dx\\y&dy\end{vmatrix}=\tfrac12(x dy - y dx) \tag{2}$$

by subtracting column $1$ to column $2$ (or plainly by looking at the figure...)

Remark : The sign of elementary area in (2) is checked positive, due to the fact that the curve is swept in the positive orientation induced by the parameterization by $\varphi$.

Answer 2

The area of the top half of the ellipse is $$\int_{-a}^a y\ dx = -ab\int_{\pi}^{0} \sin^2(\varphi)\ d\varphi={\small{\frac{1}{2}}}ab\pi$$ When applying a substitution, you need to change the limits of integration to be consistent with the original limits.