Find the area enclosed by the curve
x = t^2 − 2t, y = sqrt(t) and the y-axis.
The curve x = t^2-2t, y = √t crosses the y-axis at t=0,2
So the area is
A = ∫[0,2] xy' dt = ∫[0,2] (t^2-2t)√t dt = -8/15 √2
The area is negative because it lies to the left of the y-axis
check:
eliminating t, we have
x = y^4 - 2y^2
so the area is
∫[0,√2] y^4 - 2y^2 dy = -8/15 √2