12. Find the area enclosed by the curve x = t^2 − 2t, y = t^(1/2) and the y-axis.

Question

I'm struggling to solve this because of the lack of interval. Thanks for any help

Answer 1

Let s = t^(1/2). Then, y = s, and x = s^4 - 2 s^2.

As y = s, we may rewrite this as x = y^4 - 2 y^2. As we are finding the area enclosed by the y-axis, let us integrate with respect to y. As y = t^(1/2), we restrict y to being positive.

Then, the y-axis is defined by x = 0, so y^4 - 2 y^2 = 0 when y^2(y^2 - 2) = 0, or y = 0 and +-√2. However, as we are only considering y positive, we will be integrating from 0 to √2

As x will be negative except at the endpoints, we will take the negative of the interval..

I use I[a, b] for the integral from a to b and E[a,b] for the evaluation from a to b

Then, I[0, √2] y^4 - 2y^2 dy= y^5/5 - 2y^3/3 E[0, √2] = 4√2/5 - 4√2/3 = -8√2/15

Then, as we are taking the negative of this, our area is 8√2/15