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Finding the area between 2 curves

  1. Aug 17, 2009 #1
    1. The problem statement, all variables and given/known data

    a.) Find the area of the region bounded by the graphs of f(x) = 1/x and 2x+2y=5

    b.) Also, use the shell method to setup the integral that represents the volume of the solid formed by revolving the region bounded by the two same graphs about the y= 1/2. (Do not evaluate the integral)


    2. Relevant equations

    For area between two curves : Integral[f(x) - g(x)] dx (depending on which curve is on top)

    For shell method : 2pi Integral[ p(x) h(x) ] dx

    3. The attempt at a solution

    a.) What I did was take the second equation (2x+2y=5) and solved for y to make (y = (5-2x)/2). I then made Integral[ ((5-2x)/2) - (1/x) ] dx. For [a,b], I just used the zoom function on my calculator and got a = 1/2 and b = 2. (I'm not sure if it is correct or not) After integrating, I got 1/2[5x-x2-ln|x|]

    b.) I haven't been able to get this setup to the way that I think is correct. My answer is
    2pi Integral[ ( (5-2y)/2 ) * (2- (1/y)) ] dy . (with a = 1/2 and b = 2). I think my [a,b] are wrong because this is in terms of y, not x, since we are revolving around the y-axis.

    Also, without using Mathematica, how can I use the Integrand symbols to make this look more presentable?

    Thanks,
    Jon
     
    Last edited: Aug 17, 2009
  2. jcsd
  3. Aug 17, 2009 #2

    CompuChip

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    Science Advisor
    Homework Helper

    For a) it looks right. By the looks of it, you are also right about the intersection points, although it would be neater if you proved that (set 1/x equal to (5 - 2x)/2 and solve for x) or at least say that you read off 1/2 and 2 and plug them into 1/x and (5 - 2x)/2 to show that they are indeed intersection points.

    For b), you might find this link helpful.
     
  4. Aug 19, 2009 #3
    For part (a) i got 1/2[5x-x^2]-ln|x|
    For part (b) to revolve around y-axis you need to turn the equation into x^2 = f(y)
    and since you are revolving around y=1/2, you need to substitute x^2 = f(y-1/2).
    Hope that helps
     
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