How to get sign right for parametric area for circle For finding the area of a circle of radius $r$ using $x=r \cos(t)$ and $y=r \sin(t)$ with $t$ from $0$ to $2 \pi$, it seems that the formula usually given for parametric area, which is $\int_a^b y(t)x'(t)dt$, gives a negative result and all the examples I could find use the formula of $\int_a^b x(t)y'(t)dt$
From what I've read, for the area formula to work, you'd need a closed curve but the sign depends on the direction? From what I read on brilliant (the above paragraph, as well as the circle example), my negative sign is due to the using the first formula and not the second... however, if I use my formula with $\int_{2\pi}^{0} r \sin(x) \cdot -1 r \sin(x) \ dx$ then it works out well and I get $\pi r^2$ with the positive sign
Then reading this answer to a user with a similar issue, the suggestion is due to the sign of the product of the terms inside more than anything.
Which is the right formula to use, and is there a good way to predict which one would give the correct sign?
 A: Both formulas are correct; care needs to taken though in determining the integration limits of the parametrized variable.
In the case of $\int y(t)x'(t)dt=\int y (x)dx$, the integration is over the range $x\in [-r,r]$, which corresponds to $ [\pi,0]$ under the parameterization $x=r\cos t$ (i.e. $-r= r\cos\pi$ and $r=r\cos 0$) and the area integral for the circle is given by
\begin{align}
Area & =\int_{-r}^r [y-(-y)]\>dx = 2\int_{-r}^r y\>dx\\
&= 2 \int_\pi^0 y(t)x’(t) \>dt\\
&= 2 r^2 \int_\pi^0(-\sin^2t)\>dt\\
&= \pi r^2
\end{align}
A: For a length $l$ between points $A$ and $B$ of a curve we have $AB^2=\Delta x^2+\Delta y^2$. As $\Delta x, \Delta y \to 0$, $AB \to l$ and we have $(dl)^2=(dx)^2+(dy)^2.$ So then $(dl)^2=\left(1+(\frac{dy}{dx})^2\right)dx^2\Leftrightarrow dl=\sqrt{1+(\frac{dy}{dx})^2}dx$. The equation of the circle is of course $x^2+y^2=r^2.$ So we have $2xdx+2ydy=0\Leftrightarrow \displaystyle\frac{dy}{dx}=-\frac{x}{y}.$ Let $l$ be the circle's circumference. You have $l = \displaystyle\int_{0}^{R}{\sqrt{1+\left(\frac{x}{y}\right)^2}dx}=\int_{0}^{R}{\sqrt{\frac{R^2}{R^2-x^2}}}dx=\int_{0}^{R}{\frac{1}{\sqrt{1-\displaystyle\frac{x^2}{R^2}}}}dx=R\int_{0}^{R}{\frac{1}{\sqrt{1-\displaystyle\frac{x^2}{R^2}}}}d\left(\frac{x}{R}\right)= \\ 
=R\cdot arcsin\left(\displaystyle\frac{x}{R}\right)\Biggr|_{0}^{R}=2\pi R.$ So now integrating it is easily seen that the area of a circle is $\pi R^2.$ I know it isn't quite what you wanted but I thought it was interesting to consider.
But now of course Green's theorem comes in handy when you need to integrate parametric curves. The answer above provides a very nice example and shows how careful you need to be choosing the limits. Notice here that $t \in [0,2\pi]$ and the $x(t), y(t)$ describe parametric curves in $\mathbb{R}^2$. So obviously you can have $\displaystyle\int{y\hspace{0.11cm}dx}=\int_{0}^{2\pi}{y(t)x'(t)dt}=r^2 \int_{0}^{2\pi}{cos^2(t)dt}=4r^2\int_{0}^{\frac{\pi}{2}}{cos^2(t)dt}=I \hspace{0.5cm} (1)$, which is always an incredibly useful trick in dealing with curves... Let's solve for $I$! Now let $\theta=\frac{\pi}{2}-t\Leftrightarrow du = -dt$. So $I = 4r^2\displaystyle\int_{0}^{\frac{\pi}{2}}{sin^2(t)dt}\hspace{0.5cm} (2)$. Now $(1)+(2)=2I=4r^2\displaystyle\int_{0}^{\frac{\pi}{2}}{1d\theta}\Leftrightarrow 2I = 2\pi r^2\Leftrightarrow I = \pi r^2.$ As desired in the first place.
