- #1
vande060
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Homework Statement
I don't 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 = t2 - 2t y = t1/2 around the y axis
Homework Equations
A = ∫ xdy
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 21/2
I feel like my bound of integration should be from 0 to 21/2, but I get a negative area, so I probably should reverse the bounds, but I can't rationalize the reversal.
A = from 0 to 21/2 ∫ (t2 - 2t)t1/2 dt
= from 0 to 21/2 ( 4/5 *t5/2 - 4/3 * t3/2)
here i get 1.86 - 2.2, which is negative
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