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Homework Help: Area between curves

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the area of the bounded region enclosed by the curves: 6x+y^2=13, x=2y


    2. Relevant equations
    Integration


    3. The attempt at a solution
    Finding the area shouldn't be too much of a problem. In this particular problem, integrating with respect to y is the better choice compared to integrating with x. If you graph out the two lines, you will see that most of the area will lie under the x axis.

    My real question is do I treat that intended area as a positive value even though its under the curve? I mean the question states find the area. Would I treat this area as negative are because its under the x axis? I don't know whether to treat it positive or negative :confused:.

    Thanks for the response! :smile:
     
  2. jcsd
  3. Feb 3, 2010 #2

    Mark44

    Staff: Mentor

    Area is never negative. For example,
    [tex]\int_0^{2\pi} sin x dx = 0[/tex]
    but the area between the curve y = sin x and the x-axis, between 0 and 2pi, is 4.

    For your real question, it all boils down to how you set up your incremental area element, [itex]\Delta A[/itex]. For horizontal area elements, the area of each is (xright - xleft)[itex]\Delta y[/itex], which always gives you a positive value.

    For vertical area elements, the areaof each is (ytop - ybottom)[itex]\Delta x[/itex], which also always gives you a positive value.
     
  4. Feb 3, 2010 #3
    So any area I want can be considered to be positive all the time? Even in my example? http://www.wolframalpha.com/input/?i=plot+6x%2By^2%3D13%2C+x%3D2y+
     
  5. Feb 3, 2010 #4

    Mark44

    Staff: Mentor

    It doesn't matter whether the region is above or below the x-axis or mixed. You calculate area in such a way that it comes out positive or possibly zero.
     
  6. Feb 3, 2010 #5
    Ok. I just want to make sure. I know that integrating under the x axis returns a negative number by nature. I should be able to solve this.
     
  7. Feb 3, 2010 #6

    Mark44

    Staff: Mentor

    Not necessarily. If you set up the integral with the right area element, an integral that represents the area will come out positive. See what I said in post #2.
     
  8. Feb 4, 2010 #7
    I did out the problem and I would like someone to confirm my answer please. I got 686/9 for an answer. The bounds I used were b=1 and a=-13. If you want to see my work in more detail that's fine (I'll put it through a scanner) but i just want to make sure i have the right answer.

    Thanks!!
     
  9. Feb 5, 2010 #8

    Mark44

    Staff: Mentor

    That's correct.
     
  10. Feb 5, 2010 #9
    Thanks for your help Mark44! I thought I may have had it right but as anyone knows, all it takes is a flip in one number or sign before everything comes crashing down!

    Thanks!:smile:
     
  11. Feb 5, 2010 #10

    Mark44

    Staff: Mentor

    Yes, and I found this to be a little tricky to evaluate at -13. Had to go back and check my work a couple of times before I caught an error I had made.
     
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