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Volume of solid or revolution

  1. Feb 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid of revolution obtained by rotating the area bounded by the curves about the line indicated.
    y=x2-2, y=0 about y=-1. Need only consider part above y=-1

    2. Relevant equations
    V=∏ab[f(x)]2dx


    3. The attempt at a solution
    I'm mainly unsure of my solution, as it gives me an answer but I feel that my bounds aren't selected properly.
    Roots: -√2 to √2
    V=∏-√2 √2[(x-1)2-(-1)2]dx
    I get an answer of 21.91 u3, but as I'm working in the negative y I feel like the y upper and y lower I've selected aren't correct as the parabola is not above the line. Should I integrate in terms of x instead?
     
  2. jcsd
  3. Feb 22, 2014 #2

    haruspex

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    Did you sketch the region? Looks to me like a horizontal cylinder, axis y=-1, with a dimple in the right hand end (positive x end). That means the bounds for y depend on x. Split the range of x into two integrals.
     
  4. Feb 22, 2014 #3

    LCKurtz

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    Do you mean should you integrate in terms of y instead? I would say a ##dy## integral would be easier. Use shells. As Haruspex suggests, draw a picture if you haven't already.
     
  5. Feb 24, 2014 #4
    So, if integrating in terms of y, I would integrate x=sqrt(y+2) from -1 to 0?, So my final integral is pi (integral from -1 to 0) (sqrt(y+2))^2 dy?
     
  6. Feb 24, 2014 #5

    haruspex

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    No. This is the cylinder method. Each cylinder has length x, but what is its surface area?
     
  7. Feb 24, 2014 #6
    We didn't use terms like by shells and by cylinders in lecture so I'm trying to follow along, but since if I'm doing by cylinders would I then require the constant to be 2pi?
     
  8. Feb 24, 2014 #7

    haruspex

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    Yes, there's a 2π in there, but that's not the only error in your integral.
    Your cylindrical element has y=-1 as axis. It is dy thick. What are its length and circumference?
     
  9. Feb 25, 2014 #8
    It's radius is 1, so it's circumference is 2pi. It's length is 2x, so 2sqrt(y+2)?
     
  10. Feb 25, 2014 #9

    LCKurtz

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    That's right for the length of the ##dy## element. But it is at a variable y-coordinate ##y## and it is being rotated about ##y=-1## so its radius of rotation isn't ##1## and its circumference isn't ##2\pi##. It depends on ##y##.
     
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