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Volumes of revolution

  1. Mar 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Calculate the volume obtained by rotating the triangle bounded by y = 0, y = x, and y = 2 - x, about the line x = -2. You may use either horizontal or vertical rectangles.

    3. The attempt at a solution

    So since this is a triangle, I tried to split up the volume down to two integrals, one from 0 - 1, and the other from 1-2. I used vertical rectangles by the way. So I set up my first integral as

    [itex]2 \cdot \pi \int_{0}^{1} (x+2)(x) \,dx[/itex]
    since the shell radius would be a distance x + another 2, and then the shell height would be the x.
    My second integral is

    [itex]2 \cdot \pi \int_{1}^{2} (x+2)(2-x) \,dx[/itex]

    then I added these together, but I did not get the correct answer, could anyone show me what the error in my steps are?
  2. jcsd
  3. Mar 19, 2014 #2


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    I am issing 2). relevant equations. Where did you get the idea to set up the integrals as you did ?
    Did you make a drawing to see if you want to integrate over dx or over dy ?

    [edit]Ok, you are using vertical shells. What is their height ? x seems strange: in my drawing height decreases with x.

    [edit][edit]vertical shells require bounds in x, not in y.
  4. Mar 19, 2014 #3
    y=x is the first height,no? from 0-1 if not, then what is it
  5. Mar 19, 2014 #4


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    Ah, I see. You want to integrate over x (i.e. horizontal rings), the bounds are x=0 at the lower en x=x at the upper end , so the height is x for the first integral. And also for the second integral.

    Or do you want to use vertical rings and integrate over x ? x does not exceed 1, right ?

    Can you show your drawing ?
  6. Mar 19, 2014 #5
    I vertical cylinders, I don't think you can have vertical rings. You can have horizontal rings. Here is the graph 105v0b8.jpg
    Last edited: Mar 19, 2014
  7. Mar 19, 2014 #6


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    Your integrals are set up correctly. You must have some mistake in your evaluating them.
  8. Mar 19, 2014 #7
    Yea, I think I did miss a factor of (1/3) somewhere when I solved my integral. Thanks for the help!
  9. Mar 19, 2014 #8


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    Oh boy, my mistake: I revolved around x= -2 and had line x=0 as boundary instead of y=0. Scuse me! Eating humble pie. At least it explains why communicating was difficult.

    And yes, integrals are set up correctly -- from the start :redface:.
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