Integral Calculus: Plane Areas in Rectangular Coordinates Find the area bounded by the given curves:
$y^2+2x-2y-3=0$ and the $y$-axis
(using horizontal & vertical strip)

Finding the points:
$x = 0$ so
$$ y^2+2(0)-2y-3=0 $$
$$ y^2-2y-3=0 $$
$$ (y-3)(y+1)=0 $$
Thus points $P$ are $(0,3),(0,-1)$.
By completing the square
$y^2-2y+1=-2x+3+1$.
$$ (y-1)^2=-2(2x-2) $$
$ y=1, x=2 ,
V = (2,1) $
The parabola opens left.
