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

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The volume of the tetrahedron with vertices at (0,0,0), (1,0,0), (0,1,0), and (0,0,1) is confirmed to be 1/6. This is derived from the formula for the volume of a tetrahedron, which is (abc)/6, where a, b, and c are the lengths of the edges meeting at one vertex. The initial confusion regarding the volume being 1/2 was clarified, emphasizing that the shape is indeed a tetrahedron and not a cube. The discussion highlights the importance of correctly identifying the geometric shape to apply the appropriate volume calculation. The final consensus affirms that the calculated volume of 1/6 is accurate.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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