MHB Volume of a triangle type shape with a square bottom

[Dustinsfl]

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.

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[MarkFL]

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:

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

 

[Dustinsfl]

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:

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

How did you derive this formula? Is the 1 the volume of the cube or is that part of the triangular shape?

 

[MarkFL]

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:

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

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

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

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

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