Calculate Volume of Tetrahedron with Given Vertices | Step-by-Step Solution

madachi
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Homework Statement



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

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

Thanks for correcting it so quickly.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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