Find the area enclosed by the curve (parametric equation)

  1. 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. jcsd
  3. SammyS

    SammyS 8,270
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Have you looked at the graph?
     
  4. sure it looks like this approx
    [​IMG]

    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
     
  5. 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.
     
  6. that clears everything up thank you,
     
  7. cheers
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?