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 two curves

  1. Apr 17, 2005 #1
    Find the area of the region bounded by the curves

    y=x , y=1/(x^2) , and x=2

    I know after you sketch it you have to take the intergral of the top function minus the bottom function from the points that they intersect. I am stuck however because one function does not appear to be on top throughout the interval. Any help would be great...
     
  2. jcsd
  3. Apr 17, 2005 #2
    I'm thinking that since there was no other bound (you can't assume the y axis is a bound), I imagine their point of intersection is meant to be taken as the other bound for the function.
     
  4. Apr 17, 2005 #3

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    The attached plot is the area. It is:

    [tex]\iint_R dA[/tex]

    Right?
     

    Attached Files:

  5. Apr 17, 2005 #4

    cronxeh

    User Avatar
    Gold Member

    and what about area between x=0 and x=1?
     
  6. Apr 17, 2005 #5

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    Ok, I see what you mean. Perhaps the problem is not well defined.
     
  7. Apr 17, 2005 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The region is bounded by y= x, y= 1/x2, and x= 2.
    The only difficulty with that is that there is no lower boundary. I bet the problem also gives y= 0 as a boundary. In that case:

    Of course, y= x and y= 1/x2 cross at (1, 1). That is, the region is bounded above by y= x from x=0 to x= 1, then by 1/x2 from x= 1 to x= 2.

    The simplest way to do this is treat it as two separate regions:
    1) the region bounded by y= x, y= 0, x= 1. In fact, that's a triangle with area 1/2.

    2) the region bounded by x= 1, y= 1/x2, y= 0, x= 2.
    Integrate y= 1/x2= x-2 from x= 1 to x= 2.

    Add the areas of the two regions.
     
  8. Apr 17, 2005 #7
    I think this is a Calc I question, so the poster wouldn't have encountered double integrals yet.
     
  9. Apr 17, 2005 #8
    Possible answer?

    I got an answer of 1? can anyone verify...??
     
  10. Apr 17, 2005 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    How did you get 1? In fact, can you tell me what the bottom curve of your region is?
     
  11. Apr 17, 2005 #10
    answer of 1

    I got the answer of one assuming the bottom is y=0...i have to ask my teacher tomorrow for sure if that is suppose to be one of the curves..
     
  12. Apr 17, 2005 #11

    cronxeh

    User Avatar
    Gold Member

    i dont know how many curves you people see, but there were (from top of my head - i plotted this in the morning) 4 different regions. you have one between x=0 and x=1 - the one where 1/x^2 reaches to infinity, area is approximately 144+, then between x=1 and x=2 there is a triangular area and a parabolic area over on top.

    so this question is not very well defined. you cant really find an area 'between' curves per se- as there are interlapping curves
     
  13. Apr 18, 2005 #12
    I'm 99% sure the region is meant to be this one (in really crappy, fast sketching)...
     

    Attached Files:

  14. Apr 18, 2005 #13
    the intersection of y=1/(x^2) and y=x is at x=1

    so the area is the area under the top curve - area under the bottom curve.

    A = [I(1 to 2)(x dx)] - [I(1 to 2)((dx)/(x^2))]

    I = integral sign
     
  15. Apr 18, 2005 #14

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, or course! to both Moo of Doom and IntuitioN. I didn't draw my graph clear enough!

    You could also just integrate the difference of the y-values
    [tex]A= \int_1^2 (x- x^{-2})dx[/tex]
    (that's exactly the same as what IntuitionN gave).
     
  16. Apr 22, 2005 #15
    hey,
    its obvious from a book i read u equate x and 1/x^2,getting x as 1.this ends up being the lower boundary number,and then integrate definitely the multiple of x and 1/x^2gettiung the integral of 1/x,which becomes log [x].taking the definite step, the answer is log2,in base e.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?