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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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: , where is the area of the cross-section.
Thanks, but I'm not sure how to find A(z)
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 The length in y is equal to The area of the triangular cross-section is .
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!
Last edited by zhupolongjoe; Feb 10th 2010 at 06:39 PM.
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