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Calculus area between two curves

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data

    How to Find the area of ​​the shaded region in the image below:

    http://i47.tinypic.com/263ed5d.jpg (without space)


    Hi, please help me with this:

    How to find the common points in between the curves? and the area??

    2. Relevant equations

    y= x^2
    y= (x-2)^(1/2)
    x = 0

    3. The attempt at a solution

    I try to find the common point doing:

    x^2 = (x-2)^(1/2) but I can't do it...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 11, 2012
  2. jcsd
  3. May 11, 2012 #2
    Simple. Split it into two halves, each of which you can calculate with a simple integral.
     
  4. May 11, 2012 #3

    sharks

    User Avatar
    Gold Member

    dako, you can simply edit your post and attach your picture directly without having to use an external image host. I've made the attachment in this post.

    There is no need to split the region, as the latter is horizontally simple.
    [tex]\int \int \,dxdy[/tex]
     

    Attached Files:

  5. May 11, 2012 #4
    Hi..!! thanks!! but how may I find the limits to evaluate the integrals?
     
  6. May 11, 2012 #5

    sharks

    User Avatar
    Gold Member

    Describe the region:
    For y fixed, x varies from x= to x=
    y varies from y= to y=
     
  7. May 11, 2012 #6
    If you want to use a double integral, I suggest you learn a bit about multivariable calculus. Basically, if I'm understanding correctly, what sharks is saying is (in singlevariable students' terms), is that if we integrate x with respect to y, we'll get the same result. That is, take horizontal slices instead of vertical ones in the "infinite rectangles" definition of the Integral.
     
  8. May 11, 2012 #7
    Set the lines equal to each other to find the point of intersection. Then, integrate from 0 to that point plus from that point to the x-intercept of sqrt(2-x).
     
  9. May 12, 2012 #8
    This is the solution:

    Using horizontal
    1 Doing x= f(y)

    When y=√(2-x) , x=2-y^2
    When y= x^2 , x =√y

    The point in common is (1,1), we must evaluate from 0 to 1 this integral:

    ∫ 2-y^2 - √y dy

    It must be 1.

    Thanks!!
     
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