Find the area enclosed by the given ellipse:
$$
(x,y)=(a \cos t, b \sin t) \: , \quad 0\leq t < 2\pi
$$
I have tried to google this as well as look in my notes but I don't know where to start. please point me in the right direction.
[Math] find the area enclosed by the given ellipse
areacalculus
Related Solutions
Under a little bit more general settings, a line integral of a vector field $(p,q)$ along some oriented planar curve $L$ has the following form. $$I=\int_L\left(pdx+qdy\right).$$ Denote $\mathbf{r}=(x,y)$ and let $L$ be parameterized as $t\mapsto\mathbf{r}(t)$, $t\in[0,1]$. We can always choose another parameter $s$, the so called arc length parameter, which satisfies $\frac{d\mathbf{r}}{ds}\equiv 1$, or $ds=|\mathbf{r}'(t)|dt$. Follow your notations, $\frac{ds}{dt}(t)=v(t)$. The unit normal vector field $\mathbf{n}$ can be represented as $$\mathbf{n}=(\frac{dy}{ds},-\frac{dx}{ds})=v(t)^{-1}(\frac{dy}{dt},-\frac{dx}{dt}).$$
If we choose $\mathbf{f}=(q,-p)$ as a vector field, then the line integral above can be rewritten as $$I=\int\mathbf{f}\cdot\mathbf{n}ds=\int\mathbf{f}\cdot(\frac{dy}{dt},-\frac{dx}{dt})dt.$$
Now we know that when $L$ is a simple closed curve, and if we choose $\mathbf{f}=\frac{1}{2}\mathbf{r}$, the line integral above gives the area bounded by $L$, i.e. $$area=\frac{1}{2}\int\mathbf{r}\cdot\mathbf{n}ds=\frac{1}{2}\int(x\frac{dy}{dt}-y\frac{dx}{dt})dt=\frac{1}{2}\int_L( xdy-ydx).$$
Let's do an area preserving ($\det(T)=1$) change of coordinates, shall we:
We want a transformation which is something like $(x,y) \to (ax,by)$. If we had such a transformation, then, plugging in, we'd get something like: $b^2a^2x^2 + a^2b^2y^2=a^2b^2 \iff x^2+y^2=1$ which is just the equation of a circle.
So our matrix should be $\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}$ because $$\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} ax \\ by \end{bmatrix}$$
Notice, though, that this is not an area-preserving transformation unless $ab$ happens to equal $1$. However, this one is:
$$\left(\frac 1{\sqrt{ab}}\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}\right)\begin{bmatrix} x \\ y \end{bmatrix} = \frac 1{\sqrt{ab}}\begin{bmatrix} ax \\ by \end{bmatrix}$$
So doing the change of coordinates $(x,y) \to (\sqrt{\frac ab}x, \sqrt{\frac ba}y)$ won't change the area of your ellipse.
Plugging this in, we get $x^2 + y^2 = ab$. I assume you can find the area of this?
Best Answer
Use good old $$\int y\,dx.$$
By symmetry, you can integrate on the half-ellipse, with $t$ from $\pi$ to $0$ (i.e. $x$ from $-a$ to $a$). We have
$$\frac A2=\int_\pi^0 a\sin t\,(-b\sin t)\,dt=ab\int_0^\pi\sin^2t\,dt.$$
As$$\sin^2t=\frac{1-\cos2t}2$$ the value of the integral is $\dfrac\pi2$, giving
$$A=\pi ab.$$
Note that integration from $2\pi$ to $0$ directly yields the correct answer, as it performs a forward pass with $y$, and a backward pass, which is equivalent to a forward pass with $-y$, because $y\,dx=(-y)(-dx)$. Hence you are integrating $y_+(x)-y_-(x)$ where $y_+,y_-$ are the upper and lower arcs. This is a general method that works for closed curves. The reversal of the range, $2\pi$ to $0$, ensures clockwise rotation.