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**1. Homework Statement**

I dont really have a problem with integration here, I just need to learn how to decide what direction the integration should be in

find the area enclosed by the curve x = t

^{2}- 2t y = t

^{1/2}around the y axis

**2. Homework Equations**

A = ∫ xdy

**3. The Attempt at a Solution**

so when i plug values into the parametric equations i find that this graph comes out of the origin at y = 0 then crosses the axis again at 2

^{1/2}

I feel like my bound of integration should be from 0 to 2

^{1/2}, but I get a negative area, so I probably should reverse the bounds, but I cant rationalize the reversal.

A = from 0 to 2

^{1/2}∫ (t

^{2}- 2t)t

^{1/2}dt

= from 0 to 2

^{1/2}( 4/5 *t

^{5/2 }- 4/3 * t

^{3/2})

here i get 1.86 - 2.2, which is negative

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