Find the area enclosed by the curve   x = t2 − 3t, y =t and the y-axis.?

The answer is as follows:

(Version 1)

If x = t^2 - 3t and y = t, then the curve could also be written as x = y^2 - 3y, so the area is the integral from y = 0 to 3 of (y^2 - 3y) dy.

After integration you have (1/3)y^3 - (3/2)y^2 = 9 - 27/2 = -9/2.

The reason why the area comes out negative is that the only closed area is to the left of the y axis. The magnitude of the area is just 9/2.

(Version 2)

If x = t^2 - 3t and y = sqrt(t), then the curve can be written at

x = y^4 - 3y^2, and the area is the integral from y = 0 to sqrt(3) of

(y^4 - 3y^2) dy.

After integration you have (1/5)y^5 - y^3

= sqrt(3)*(9/5 - 3) = -6*sqrt(3)/5.

As in Version 1, the reason why the area comes out negative is that it's to the left of the y-axis; the magnitude of the area is 6*sqrt(3)/5.