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Finding the volume of a solid revolution (shell method)

  1. Aug 28, 2012 #1
    Let f(x)=9-x^2. Let A be the area enclosed by the graph y=f(x) and the region y>=0.
    Suppose A is rotated around the vertical line x=7 to form a solid revolution S.
    So, using the shell method, I was able to find the indefinite integral used.
    I found the shell radius to be (7-x) and the shell height to be (9-x^2)
    Therefore, the volume is 2∏∫(7-x)(9-x^2)dx = 2∏∫(63-9x-7x^2+x^3)Δx
    However, I am just unsure for what the interval should be. For whether it should be [-3,3] or [0,7]. So, which one is it?
    I also tried using the disc method, with π ∫ [7² - (9 - y)] dy from 0 to 9 and got the answer 801∏/2. Is that right?
    Any help is appreciated. Thanks.
     
  2. jcsd
  3. Aug 28, 2012 #2
    To find your limits, think about the description of the area A. It is the area enclosed by the function and the region y >= 0. Picture this area on a graph (make a sketch if it helps), and it should be pretty clear that the integral should go from x=-3 to x=3. Using the other method, your formulas for your radii aren't quite right. Your outer radius should be [tex]7+\sqrt{9-y}[/tex] and your inner radius should be [tex]7-\sqrt{9-y}[/tex]. Again, picture the situation on a graph (for this one, it might be especially helpful to actually draw a quick sketch) and you should see why you get that.
     
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