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  1. MHB Master

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    #1
    How do I find the volume of this shape? The bottom is a square in the xy plane where \(0\leq x,y\leq 1\).

    The object isn't a prism or pyramid so I am not sure what to do.


  2. Pessimist Singularitarian
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    #2
    If I am interpreting this correctly, for $0\le z\le1$ you have a cube whose sieds are 1 unit in length, and for $1\le z\le2$ you have a solid whose cross-sections perpendicular to either the $x$ or $y$ axes are right triangles whose bases are 1 unit in length and altitudes vary linearly from 0 to 1, and so the volume by slicing is:

    $ \displaystyle V=1+\frac{1}{2}\int_0^1 x\,dx=\frac{5}{4}$

  3. MHB Master

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    #3 Thread Author
    Quote Originally Posted by MarkFL View Post
    If I am interpreting this correctly, for $0\le z\le1$ you have a cube whose sieds are 1 unit in length, and for $1\le z\le2$ you have a solid whose cross-sections perpendicular to either the $x$ or $y$ axes are right triangles whose bases are 1 unit in length and altitudes vary linearly from 0 to 1, and so the volume by slicing is:

    $ \displaystyle V=1+\frac{1}{2}\int_0^1 x\,dx=\frac{5}{4}$
    How did you derive this formula? Is the 1 the volume of the cube or is that part of the triangular shape?

  4. Pessimist Singularitarian
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    #4
    Yes the 1 is the volume of the cubical portion of the solid, and for the upper part, the volume of a particular slice is:

    $ \displaystyle dV=\frac{1}{2}bh\,dx$

    where the base is a constant 1 and the height is $x$, hence:

    $ \displaystyle dV=\frac{1}{2}x\,dx$

    and so summing the slices (and adding in the cubical portion), we find:

    $ \displaystyle V=1+\frac{1}{2}\int_0^1 x\,dx$

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