Find the area enclosed by the curve x = t^2 - 2t , y=sqrt(t), and the y-axis.

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First find the the value of t where the curve intersects the Y-axis. This is when x = 0.

x = t^2 - 2t = 0 = t(t - 2)

So t= 0 and t = 2

dA = (0 - x)*dy .... Since the curve has negative x in this region

y = SQRT(t) and dy = [(1/2)/SQRT(t)]dt

dA = [2t - t^2][(1/2)/SQRT(t)]dt

dA = [t^(1/2) - (1/2)t^(3/2)]dt

Integrate to get: A = (2/3)t^(3/2) - (1/5)t^(5/2)

Now evaluate from t= 0 to t = 2.

Area = [(2/3)2^(3/2) - (1/5)2^(5/2)] - [0]

Area = SQRT(2)[4/3 - 4/5]

Area = SQRT(2)[8/15) = 0.754