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Homework Help: Finding the volume of these figures

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the described figures

    1. A pyramid with height h and base an equilateral triangle with side a (a tetrahedron).

    2. Find the volume common to two spheres each with radius r, if the center of each sphere lies on the surface of the other sphere.

    These are hard problems and I really do not know what to do. Our teacher gave this to us as a challenge and I really would like to know how to solve these two. I hope you can help me by explaining each important step you will take. I hope you can make it detailed as much as possible. I really am having a hard time in Calculus. Thanks! :)
     
  2. jcsd
  3. Sep 25, 2011 #2

    CompuChip

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    For 1): you can slice the tetrahedron horizontally into pieces of height dz. Then each piece will again be an equilateral triangle whose sides decrease from a at z = 0 to 0 at z = h.

    For 2): you can slice the intersection of the spheres into circles of thickness [itex]d\theta[/itex] whose radius [itex]r(\theta)[/itex] increases from [itex]r(\theta = -\pi / 2) = 0[/itex] to [itex]r(\theta = 0) = r[/itex] (see attachment) and you can use some fancy trig work to find the expression for r(theta).
     

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  4. Sep 25, 2011 #3

    HallsofIvy

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    It was you the teacher was challenging, not us! There are a number of different ways to do these problems ranging from looking up standard formulas to using Calculus as CompuChip suggests.
     
  5. Sep 25, 2011 #4
    I'm sorry but I don't exactly get it. I'm really having a hard time in Calculus right now.

    For #1, what is dz? Just a representation for height? and I can't picture properly the tetrahedron I need to slice.

    For #2, I really don't get it. I'm sorry.

    I hope you'll still explain it to me.
     
  6. Sep 25, 2011 #5
    I got the 1st one already. Only #2 left.
     
  7. Sep 26, 2011 #6

    CompuChip

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    Here is another hint, hopefully it clarifies a bit better what I meant.
    The volume of the little circular disc (actually, it's a cylinder with radius r and thickness dtheta) that I drew is [itex]\pi r^2 \, d\theta[/itex]. Of course you'll have to express r as a function of theta before you do the integration (and find the appropriate limits for theta).
     

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