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Rotation around an axis

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data
    The curve y=1/x and the line y=2.5-x enclose an area together. Determine the exact volume of the rotating body that is formed when this field rotates about
    a) The x-axis and b) The y-axis

    2. Relevant equations
    The formula for rotation around the x-axis is pi*integrate from b to a y^2 dy..
    The formula for rotation around the y-axis is pi*integrate from d to c x^2 dx.


    3. The attempt at a solution
    I'm gonna start with a). So what I've learned is that around the x-axis, first I have to find the points from where I'm gonna integrate (i.e. integrate from z to y and so on). To get these two points, I'm going to put the curve equal to the line so I know where they intersect.
    2.5-x=1/x => 2.5x-x^2=1 => x^2-2.5x+1 = 0 and I get x(1) = 2 and x(2) = 0.5.

    So if I'm going to find out the answer, I'll also use (1/x)^2 - (2.5-x)^2. So the total would be:

    pi * integrate from 2 to 0.5 for (1/x)^2 - (2.5-x)^2 which gives me the correct answer 1.125*pi.

    Okay, so how do I do it for b then?
    Take a look here: http://www.wolframalpha.com/input/?i=y=1/x,+y=2.5-x
    I want to calculate the volume for the rotation that is both under the purple and blue line. But how do I get that? It is obvious that the "integration limits" are from 2.5 to 0 from that graph. But how do I express it in a formula?

    If I do the same as earlier, (1/x) - (2.5-x) Then doesn't that area under the purple line eliminate all of it that is under the blue and then some?
     
  2. jcsd
  3. May 4, 2015 #2
    Think of it this way:
    You'll get the volume of the figure if you first find the volume under the purple line and the remove the volume [itex]between[/itex] the purple line and the blue curve
     
  4. May 4, 2015 #3
    So first I'd take the total volume under the purple line. But how do I get the volume that is between the purple line and the blue curve? Purple minus blue and then the total volume under purple minus whatever I got from the second one?
     
  5. May 4, 2015 #4
    Exactly!
     
  6. May 4, 2015 #5

    SammyS

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    In this and your other recent thread, you have the form of the integral wrong.

    For the rotation about the x-axis, y is a function of x, and the integration variable is x, not y, so that should be dx.

    The integral is: ##\displaystyle\ \pi\int_{x_1}^{x_2} y^2\, dx ##
     
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