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The Integration problem that's still giving me a headache (a week later)

  1. Mar 8, 2005 #1
    Last week (or longer) I asked how to integrate the following rotated 360 degrees about the x-axis (and I still can't figure it out):
    y= [(1-(x-3)^2)/9)]^(1/2) + 2

    When I applied the boundaries of 0 to 6, I got 23.2 cm^3, but I don't think its the boundaries that are my problem, I think it is the integration. Help! I need this for tomorrow. Thanks!!
  2. jcsd
  3. Mar 8, 2005 #2


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    First, it makes no sense to "integrate the following rotated 360 degrees"! You integrate a function but you can't rotate a function- you rotate a geometric object!

    My best guess is that you want to find the volume of the figure formed by rotating the graph of y= [1-(x-3)/9]1/2+ 2 around the x axis. Doing that, each cross section is circle of radius y so the each cross section has area πy2 and so the function you want to integrate πy2dx. That should not be hard.

    If y= [1- (x-3)2/9]1/2+ 2 then (y-2)2= 1- (x-3)2/9 so (x-3)2/9+ (y-2)2= 1. That's an ellipse with center at (3,2) and semi-axes of length 1 (in the y direction) and 3 (in the x direction). It doesn't cross the x-axis but since you specify y= [1- (x-3)2/9]1/2+ 2, I GUESS you mean to drop straight lines from the largest and smallest x values, 3-3= 0 and 3+3= 6, the values you mention so I guess I am interpreting this correctly. If that's the case then your integral is
    [tex]\pi\int_2^4 y^2 dx= \pi\int_2^4(1-\frac{(x-3)^2}{9}+ 4(1-\frac{(x-3)^2}{9})^{1/2}+ 4)dx[/tex]
    You should be able to integrate all except that middle term, with the 1/2 power, easily.

    To integrate [itex]4(1- \frac{(x-3)^2}{9})^{1/2}[/itex], use a trig substitution.
    let [tex]sin(\theta)= \frac{x-3}{3}[/itex]. Then [itex]cos(\theta)d\theta= \frac{1}{3}dx[/itex] so that [itex]3cos(\theta)d\theta= dx [/itex] and [itex](1-\frac{(x-3)^2}{9})^{1/2}= (1- sin^2(\theta))^{1/2}= cos(\theta)[/itex].
    Last edited: Mar 8, 2005
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