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

1. ### vande060

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

Last edited: Apr 19, 2011
2. ### SammyS

8,270
Staff Emeritus
Have you looked at the graph?

3. ### vande060

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

4. ### eczeno

242
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.

5. ### vande060

187
that clears everything up thank you,

242
cheers