Question

∭E xz dV,where E is the solid tetrahedron with vertices (0,0,0),(1,0,0),(1,1,0) and (1,0,1)

Answer 1

Anyone can confirme if the correct boudaries are:

0≤x≤1

0≤y≤x

0≤z≤x−y

Answer 2

@mathmale

Answer 3

@Loser66

So it seems that I've been right all along. The boundaries are the ones I've mentioned above. To calculate you have to check the plane xy where you get \[y \le x\]. Then you have to calculate the plane that bounds the tetrahedron in the z-axis.

For that I've used the 3 vertices of said plane A= (0,0,0), B= (1,0,1) and C= (1,1,0). Using the vertices I got 2 vectors:
AB = (1,0,1) - (0,0,0) = (1,0,1)
AC = (1,1,0) - (0,0,0) = (1,1,0)

Then I've calculated the external product of AB and AC.

AB x AC = |i j k| = -1i +1j +1k
|101|
|110|

So now you can get the plane equation: -1(x-0)+1(y-0)+1(z-0) = 0 (I'm using the vertice A, but you can use the one you like most).

Simplifying you get -x + y + z = 0 <=> z = x - y, that is the upper bound of Z.

Substituing in the integrals:

\[\int\limits_{0}^{1}\int\limits_{0}^{x}\int\limits_{0}^{x-y} xz dz dy dx\]

Solving you get the volume that is 1/30.

Answer 4

thank you for posting it.