# find the area enclosed by the curve (parametric equation)

by vande060
Tags: curve, enclosed, equation, parametric
 P: 187 1. The problem statement, all variables and given/known data 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 = t2 - 2t y = t1/2 around the y axis 2. Relevant 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 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 cant 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
 PF Patron HW Helper Sci Advisor Emeritus P: 7,076 Have you looked at the graph?
 P: 187 sure it looks like this approx starting at t=0 there the curve is at the origin, and at point t = 2 the graph it at the point 0,21/2 so the bounds i used seemed right
P: 242

## find the area enclosed by the curve (parametric equation)

the area should be negative; it is 'below' the y-axis.

there is one mistake that i can see. if y = t^(1/2), then dy = (1/2)*t^(-1/2) dt. you seem to have integrated xy instead of xdy.
P: 187
 Quote by eczeno the area should be negative; it is 'below' the y-axis. there is one mistake that i can see. if y = t^(1/2), then dy = (1/2)*t^(-1/2) dt. you seem to have integrated xy instead of xdy.
that clears everything up thank you,
 P: 242 cheers

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