1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area between a line and a parabola

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the area enclosed by the line y = x-1 and the parabola y^2 = 2x+6



    2. Relevant equations I don't think there are any special formulas or anything



    3. The attempt at a solution

    Well, I graphed the two given equations as √2x+6 and x-1 and got intersects at point (-1,-2) and (5,4).

    A = ∫(√2x+6) - (-√2x+6)dx
    A = 2∫(√2x+6)dx

    then I used u substitution to get 2/3u^(3/2). I turned the u back to 2x+6 and then solved it for the limits -2 and -1 (Solved 2/3(2x+6)^3/2))

    but then the numbers get weird and become really long decimals...which I doubt I should be getting. Where did I go wrong? :(
     
  2. jcsd
  3. Feb 8, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi VWhitehawk! Welcome to PF! :wink:
    Have you tried looking at it sideways, and integrating wrt y instead of x ? :smile:
     
  4. Feb 8, 2012 #3

    Mark44

    Staff: Mentor

    Your integrand is incorrect. The left boundary is the parabola, and the right boundary is the line, which you seem to have neglected. If you use horizontal strips, you can use one integral, but if you use vertical strips (as you have done), you need two separate integrals.

     
  5. Feb 8, 2012 #4
    Oh, so the it should look like "∫((√2x+6) - (x-1))dx to take the line into account?

    Not sure what you mean by doing it vertically, but from the way you say it, it sounds easier than doing it the horizontal way! lol
     
  6. Feb 8, 2012 #5

    Mark44

    Staff: Mentor

    Here you are using vertical strips of width Δx. This won't work, since some strips run between the upper and lower parts of the parabola, while other strips run between the upper part of the parabola and the line. Since the strips are different, you need two different integrals.

    Have you actually sketched a graph of the region?
    Horizontal strips is width Δy are much simpler.
     
  7. Feb 8, 2012 #6
    I have drawn it and the little rectangle too. I flipped the drawing sideways like you said but I just don't get what is supposed to be my integrand.
     
  8. Feb 8, 2012 #7

    Mark44

    Staff: Mentor

    What little rectangle?

    And what do you mean when you say you flipped the drawing sideways? tiny-tim was suggesting that you use horizontal strips, not that you turn the drawing sideways.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Area between a line and a parabola
  1. Area between two lines (Replies: 5)

Loading...