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Area of simple curve bounded by

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the area of the curve 2/sqrt(x) bounded by x = 0, y = 3, y = 1

    2. Relevant equations

    The textbook claims the answer is 3.

    3. The attempt at a solution

    I tried both vertical and horizontal elements, but got different answers than 3.

    Here's my attempt at vertical elements:

    since y = 1 is the farthest right value of x, I solve 1 = 2/sqrt(x) for X to find x upper, which is 4.

    Now I form my integral: integral(4-0) of 2/sqrt(x) (dx)

    Integrate: 4 * sqrt(x)

    Solve the definite integral: (4 * sqrt(4)) - (4 * sqrt(0)) = 8

    What have I done wrong?
  2. jcsd
  3. Nov 16, 2008 #2


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    Staff Emeritus
    Science Advisor

    When y= 3, x= 4/9. But your left boundary is x= 0 so you have the lines y= 1 and y= 3 as upper and lower bounds until x= 4/9. The area is
    [tex]\int_0^{4/9} (3-1)dx+ \int_{4/9}^4 2/\sqrt{x}dx[/tex]

    What did you get when you integrated with respect to y?
  4. Nov 16, 2008 #3
    Thanks I finally understand why I have to use two integrals to figure out this question if I use verticle elements.

    For horizontal elements, I did this:

    Integral(3-1): (4/y^2) (dy)

    Integrate: -4 * y^(-1)

    Solve: -4*(1/3) - (-4)(1)

    Equals: 2 and 2/3

    What did I do wrong this time?
  5. Nov 16, 2008 #4


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    Staff Emeritus
    Science Advisor

    Nothing. Unless you have stated the problem incorrectly, that is the correct answer.
  6. Nov 16, 2008 #5
    My textbook is incorrect as I suspected. It claims the answer is 3.

    Thanks for your help.
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