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Intersection between the solid and a plane perpendicular

  1. Jul 4, 2010 #1
    1. The problem statement, all variables and given/known data
    A solid has a circular base of radius r=4. Find in each case the volume of the solid if all intersection between the solid and a plane perpendicular to a set diameter is:

    a) A square
    b) A triangle rectangle isosceles which one of its sides is on the circular base.

    3. The attempt at a solution

    The first I thought was in some kind of cylindroid. I know I must find an equation that I can integrate. So, the sides of the squares will be given by:

    [tex]z=2\sqrt{r^2-x^2}[/tex]

    Is the integration given by: [tex]\displaystyle\int_{-4}^{4}[2\sqrt{r^2-x^2}]^2dx=4\displaystyle\int_{-4}^{4}(r^2-x^2)dx[/tex] for the square?

    And how can I find the volume of the other solid b)?
     
  2. jcsd
  3. Jul 4, 2010 #2

    Dick

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    Re: Solids

    That looks fine to me. Ok, so the area of a square with side z is z^2. And you integrated the area to get the volume. Fine. What's the area of a isosceles right triangle with side z? I'm not sure why this is confusing you. Are you worried about Odysseus' return?
     
    Last edited: Jul 4, 2010
  4. Jul 5, 2010 #3
    Re: Solids

    Haha I still drunk from Blooms day :p

    The problem I got is... the area of the triangle: [tex]\diplaystyle\frac{bh}{2}[/tex]
    How do I get h? Pythagoras?
     
    Last edited: Jul 5, 2010
  5. Jul 5, 2010 #4

    Dick

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    Re: Solids

    In the triangle I'm thinking of both b and h are z. In an isosceles right triangle both sides are equal, aren't they?
     
  6. Jul 5, 2010 #5
    Re: Solids

    If both of them are equal one each other, then they are not equal to the base. So I must find h....
     
  7. Jul 5, 2010 #6

    Dick

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    Re: Solids

    There are two different ways to draw the picture. The question says 'one of its sides is on the base'. I took this to mean that it isn't the hypotenuse that has length z, but rather one of the legs. I'm probably wrong. If I am, yes, use pythagoras to find the missing length.
     
  8. Jul 5, 2010 #7
    Re: Solids

    Sorry, I didn't give a righteous interpretation at what you said in the first place. You're right, it's easier to think of both, the base and the high as the information I already get to. So, it won't be hard to find the volume from there.

    Thanks Dick.
     
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