Volume of tetrahedron

1. Apr 22, 2010

1. The problem statement, all variables and given/known data

Find the volume of the tetrahedron with vertices at $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$

3. The attempt at a solution

I worked out the triple integral and found out that the volume is $\frac{1}{6}$? Is this correct? I know there is probably a much quicker way working the volume by just using the volume formula for tetrahedron. However, I am not sure which value to substitute to the formula, so could you just tell me whether this answer is right or not?

Thanks!

2. Apr 22, 2010

Mr.Miyagi

The vertices you give do not make a tetrahedron. Try to draw the points in three dimensions. You'll see the volume is just half that of a cube with length 1. So the volume you're seeking should be 1/2.

3. Apr 22, 2010

Dick

For a tetrahedron, like a cone, the area is (1/3)*(area of the base)*height. So (1/3)*(1/2)*1=1/6, yes.

4. Apr 22, 2010

HallsofIvy

Staff Emeritus
Mr. Miyagi is wrong. That is in fact a tetrahedron, it is 1/6 of a cube, not 1/2, and its volume is, indeed, 1/6.

More generally, the volume of the tetrahedron is with vertices at (0, 0, 0), (a, 0, 0), (0, b, 0), and (0, 0, c) is (abc)/6.

5. Apr 22, 2010

Mr.Miyagi

Ugh, sorry about that... Is it too late to claim temporary insanity? :uhh:

Thanks for correcting it so quickly.