Let's say we have a given parametric function $x=x(t)$, $y=y(t)$, where $x,y$ are continous on some segment $[t_1,t_2]$. From what I've seen there are multiple ways to find the area enclosed by the curve (if the curve is closed) of these function:
$(1)$ $$A_1=\int_{t_1}^{t_2}{y(t)x'(t)dt}$$ when $x'(t)\ge 0$ and $y(t)\ge0$
$(2)$ $$A_2=\int_{t_1}^{t_2}{x(t)y'(t)dt}$$ when $x(t)\ge 0$ and $y'(t)\ge0$
$(3)$ $$A_3=\frac{1}{2}\int_{t_1}^{t_2}{(x(t)y'(t)-y(t)x'(t))dt}$$
I have a few questions regarding this.
- If all conditions of $(1)$ and $(2)$ are satisfied will $A_1=A_2$?
- When do we use $(3)$ and what are the conditions that need to be satisfied?
- If we were, for example, to use $(1)$ or $(3)$ instead of $(2)$ (when $(1)$ or $(3)$ wouldn't yield the correct solution), how would the result differ? (if possible with an example)
The reason all these expressions yield (up to sign) the are encloseed by your closed curve $\gamma$ comes from Stokes theorem:
If $\gamma:[t_1,t_2]\to\mathbb R^2, \gamma(t) = (x(t),y(t))$ is a closed positively oriented differentiable curve enclosing a region $\Omega\subset\mathbb R^2$, then for any differentiable vector field $\vec F=(F_1,F_2):\mathbb R^2\to\mathbb R^2$ one has $$ \iint_\Omega (\partial_x F_2(x,y)-\partial_yF_1(x,y))\,\mathrm dx\,\mathrm dy = \int_{t_1}^{t_2}\langle (\vec F\circ \gamma)(t),\dot\gamma(t)\rangle\,\mathrm dt. $$ Now if the integrand in the expression on the left hand-side is constant with value 1, the expression yields the area of $\Omega$. There is infinitely many choices for this to happen. Take for instance $\vec F(x,y) = (0,x)$. In this case the right hand-side becomes $$ \int_{t_1}^{t_2}x(t)\dot y(t)\,\mathrm dt. $$ If one choses $\vec F(x,y) = (-y,0)$ instead, one obtains $$ -\int_{t_1}^{t_2}y(t)\dot x(t)\,\mathrm dt. $$ and the two expressions can be seen to be equivalent using integration by parts. Another choice is $\vec F(x,y) = \frac{1}{2}(-y,x)$ which gives $$ \frac{1}{2}\int_{t_1}^{t_2}(x(t)\dot y(t)-y(t)\dot x(t))\,\mathrm dt $$ so there is quite some flexibility in choosing an integrand which will yield the area of $\Omega$ when integrating along $\partial \Omega$.