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Use only double integrals to find the volume of the solid tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,2,0)$ and $(2,2,0)$.

I know you plot the points on $xyz$-plane, but how do you get the equation of the plane that goes in the equation by the four vertices?

gen-ℤ ready to perish
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user177780
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    This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant. – Alan Oct 04 '14 at 00:55
  • I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions. – gen-ℤ ready to perish Aug 28 '17 at 00:55

3 Answers3

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Integrating with respect to $x$ and $y$, then
- the "base" triangle in that plane is given by $(0,0,0)$,$(0,2,0)$,$(2,2,0)$, and $(x,y)$ shall vary in that triangle;
- the "roof" plane, that is the plane that bounds the $z$ height, will be the plane passing through $(0,0,1)$,$(0,2,0)$,$(2,2,0)$.

G Cab
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At each $z$, the tetrahedron is of a square disk with the size $2(1-z)\times 2(1-z)$, which you could also get by performing a simple integral.

So, the volume integral is over a stack of square disks form 0 to 1 in $z$ coordinate.

$$V=\int_0^1 4(1-z)^2dx= \frac{4}{3}$$

Quanto
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HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)\times (2,2,-1)$ is a normal vector for the plane you want.

Laars Helenius
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