Consider the tetrahedron with vertices

(0,0,0)

(0,1,2)

(0,2,0)

(3,0,0)

Find its volume using integration. (No double or triple integrals though)

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- Feb 10th 2010, 11:34 AMzhupolongjoeTetrahedron volume
Consider the tetrahedron with vertices

(0,0,0)

(0,1,2)

(0,2,0)

(3,0,0)

Find its volume using integration. (No double or triple integrals though) - Feb 10th 2010, 11:38 AMicemanfan
Find a formula in terms of z for the area of a cross-section of the tetrahedron by a plane parallel to the xy-plane, and then integrate:

$\displaystyle \int _0 ^2 A(z) dz$,

where $\displaystyle A(z)$ is the area of the cross-section. - Feb 10th 2010, 12:35 PMzhupolongjoe
Thanks, but I'm not sure how to find A(z)

- Feb 10th 2010, 12:43 PMicemanfan
First, observe that the dimensions of the triangle decrease linearly as you move from z = 0 to z = 2. Therefore,

The length in x is equal to $\displaystyle 3 - 1.5z$

The length in y is equal to $\displaystyle 2 - z$

The area of the triangular cross-section is $\displaystyle 0.5(3 - 1.5z)(2 - z)$. - Feb 10th 2010, 02:15 PMzhupolongjoe
Okay, I see what you did. Sorry, but the only thing I still don't understand is how you got 3-1.5z. I understand the 2-z I think, but not this part.

EDIT::Nevermind, got it!