find the area enclosed by the curve (parametric equation)


by vande060
Tags: curve, enclosed, equation, parametric
vande060
vande060 is offline
#1
Apr19-11, 03:20 PM
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
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SammyS
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#2
Apr19-11, 04:17 PM
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Have you looked at the graph?
vande060
vande060 is offline
#3
Apr19-11, 04:40 PM
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

eczeno
eczeno is offline
#4
Apr19-11, 05:59 PM
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.
vande060
vande060 is offline
#5
Apr19-11, 06:50 PM
P: 187
Quote Quote by eczeno View Post
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,
eczeno
eczeno is offline
#6
Apr19-11, 07:03 PM
P: 242
cheers


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