find the area enclosed by the given ellipse Find the area enclosed by the given ellipse:

$$
(x,y)=(a \cos t, b \sin t) \: , \quad 0\leq t &lt; 2\pi
$$

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.
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.
area is $\pi ab$ also you can derive the area using definite integration