I am assigned to calculate the area beneath the curve $y=x^2$ and above the $x$-axis using the formula
$$A=\frac{1}{2}\int_C x\,dy\,-y\,dx$$
from $0\le x\le2$
while this seems simple to me I parameterize $x=t,y=t^2$ so $dx=dt$ and $dy=2t\,dt$. Plugging this into the equation I get $$A=\frac{1}{2}\int_0^2t\,(2t)\,dt\,-t^2\,dt$$ which gets you $\frac{1}{2}\int_0^2\,t^2\,dt=\frac{4}{3}$ however, integrating regularly you get $A=\int_0^2 x^2\,dx$ which is $\frac{8}{3}$. Could anyone tell me what is going wrong?
In the expression,
$$A=\oint_C (x\,dy-y\,dx)$$
$C$ represents a closed curve, oriented counter-clockwise, that bounds the region over which the area is calculated.
In the problem of interest, we have
$$\begin{align} \oint_C x\,dy-y\,dx&=\int_0^2 (-0)\,dx+\int_0^4 (2)\,dy+\int_{-2}^0(2t^2-t^2)\,dt\\\\ &=4-\frac43\\\\ &=\frac83 \end{align}$$