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Volume of solid

  1. Dec 26, 2006 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:

    x+3=4y-y^2
    x=0
    About x-axis

    2. Relevant equations

    3. The attempt at a solution
    I tried solving x+3=4y-y^2 for y, but I can't get it even though it seems simple.

    *Edit*
    Can someone check if I set up this integral correctly (for a different problem)?
    Find volume of region bounded by given curve and rotated around given axis:
    y=x^3
    y=8
    x=0
    Rotate about x=2

    Integral from 0 to 8 of: pi(2-0)^2 - pi(2-x^1/3)^2 dy

    thanks
     
    Last edited: Dec 26, 2006
  2. jcsd
  3. Dec 26, 2006 #2

    cristo

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    y cannot be expressed as a function of x. Try plotting the graph, and you'll see that for each value of x there will be two values of y.

    Should this not be [tex] \pi\int_0^8 2^2-y^{2/3}dy [/tex]?
     
  4. Dec 26, 2006 #3

    Fredrik

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    I started with this one and right now I don't have time to do the other one. I think you can express this volume as

    [tex]\int_0^8\pi (2-y^{1/3})^2 dy[/tex]

    or equivalently as

    [tex]\int_0^2 2\pi (2-x) x^3 dx[/tex]
     
  5. Dec 26, 2006 #4

    Fredrik

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    First of all, you have to figure out what this region looks like. Define f by [itex]f(y)=-y^2+4y-3[/itex]. Calculate f´ and f". Solve the equation f´(y)=0 for y. Is f"(y) positive or negative when y is such that f´(y)=0? Find the points where the curve intersects the line x=0.

    When you've done this you will know enough about what the region looks like to proceed. You have two options, either integrate over y (which means that you add up the volumes of cylindrical shells of thickness dy) or integrate over x (which means you add up the volumes of circular discs with a hole at the center). I think it's easier to integrate over y.

    The solutions of the equation [itex]z^2+az+b[/itex] are [itex]z=-a/2\pm\sqrt{(a/2)^2-b}[/tex].
     
  6. Dec 27, 2006 #5
    Thanks for help everyone. I have another question but didn't want to create another thread. I'm having trouble with these volume questions, so I hope some more examples will help me understand them better.

    The height of a monument is 20 m. A horizontal cross-section at a distance x meters from the top is an equilateral triangle with side x/4 meters. Find the volume of the monument.
     
  7. Dec 27, 2006 #6

    Fredrik

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    Do you know how to calculate the area of an equilateral triangle with side x/4? Call that A(x). The volume of a triangular slice of thickness dx is then A(x)dx. Integrate from 0 to 20.

    [tex]\int_0^{20} A(x) dx[/tex]
     
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