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Homework Help: 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?

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


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    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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