- #1
mikfig
- 15
- 0
Homework Statement
Let points P1: (1, 3, -1), P2: (2, 1, 4), P3: (1, 3, 7), P4: (5, 0, 2)...form the vertices of a tetrahedron. Find the volume of the tetrahedron.
Homework Equations
V = 1/3 ah
A = area of base
h = height of tetrahedron
The Attempt at a Solution
I wanted to solve this using the fact that | u x v | = area of a parallelogram formed by the vectors u and v. So I designated P1 as the "top vertex" and the other three as the vertices forming the "base" of the tetrahedron.
Then I figured I would divide the tetrahedron into "triangle slices" whose area was | u x v| / 2 while u and v where the two vectors forming this slice.
Then:
u = (P3 +t(P1 - P3)) - (P2 +t(P1 - P2))
v = (P4 +t(P1 - P3)) - (P2 +t(P1 - P2))
And I integrated | u x v | / 2 from t = 0 to t = 1. After simplifying | u x v | / 2 into an expression in terms of t, I get:
1/2 sqrt(75t^4 - 290t^3 +450t^2-290t + 75)
So when I integrated 1/2 sqrt(75t^4 - 290t^3 +450t^2-290t + 75) dt from 0 to 1, I got approximately 2.25426. Note: I had to integrate numerically as Mathematica couldn't integrate analytically.
The real answer is 20/3 or 6.66667. I looked at this and saw that if you multiply my answer by 3 then you get 6.76278, which is within 0.0961158 of the real answer.
I'm not sure if that means I am missing a constant term in my integral, but that 1/3 in the tetrahedron volume equation caught my eye as I didn't understand the reason for it when I read a proof on a website I can't remember.
Anyways, now I am thinking that I overlooked the differential and that dt should have been dy in the integral, and I would have to then somehow express dt in terms of dy. But I haven't looked at that part yet. I think I will take a look at it tomorrow, after my AP Macroecon test.
Thanks a lot,
mikfig