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Math Help - volume of wedge cut from cylinder

  1. #1
    Eater of Worlds
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    volume of wedge cut from cylinder

    Here's a cool calc problem you may wish to tackle. Perhaps you've seen it before.

    "A wedge is cut from a cylinder of radius r by two planes, one perpendicular to the axis of the cylinder and the other making an angle @ with the first. Find the volume of the wegde by slicing perpendicular to the y-axis".

    Then find the volume by slicing perpendicular to the x-axis.

    Here's a diagram to help. I used @ instead of theta because I used paint to construct the diagram.
    Last edited by galactus; November 24th 2008 at 06:38 AM.
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  2. #2
    Senior Member JaneBennet's Avatar
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    From the diagram below, it looks like

    V\ =\ 2\int_0^r{x\tan{\alpha}\sqrt{r^2-x^2}}\,\mathrm{d}x

    Which will come out to be \frac{2}{3}\,r^3\tan{\alpha}.
    Attached Thumbnails Attached Thumbnails volume of wedge cut from cylinder-lookslike.gif  
    Last edited by JaneBennet; December 24th 2007 at 05:37 PM.
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  3. #3
    Eater of Worlds
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    Good Jane. That's certainly it. It appears you done it perp. to the x-axis.

    Here's the perp. to the y-axis version:

    Let the height of the triangle be h

    Then we have \frac{h}{x}=tan({\theta})

    h=xtan({\theta})

    The area of the triangle is \frac{1}{2}hx=\frac{1}{2}x^{2}tan({\theta})=\frac{  1}{2}(r^{2}-y^{2})tan({\theta})

    because x^{2}=r^{2}-y^{2}.

    And our integral is:

    V=\frac{1}{2}tan({\theta})\int_{-r}^{r}(r^{2}-y^{2})dy=\frac{2}{3}r^{3}tan({\theta})
    Last edited by galactus; December 24th 2007 at 06:31 PM.
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