One-dimensional integration - flux

[drynada]

<Moderator's note: Moved from a technical forum and thus no template.>

Calculate flux of the vector field $$F=(-y, x, z^2)$$ through the tetraeder $$T(ABCD)$$ with the corner points $$A= (\frac{3}{2}, 0, 0), B= (0, \frac{\sqrt 3}{2},0), C = (0, -\frac{\sqrt 3}{2},0), D = (\frac{1}{2},0 , \sqrt 2)$$ by one dimensional integrtion.

I calculated div, where $$divF = 2z$$

I can see from the points distribution that $$z\in [0,\sqrt 2] $$ And according to theorem of Gauss I should apply this formula.

$$\iiint_TdivFdv$$

How can I separate the integrals to get one dimensional?

 

[Charles Link]

The tetrahedron they give you has its base it the x-y plane. What is the area of the base? Can you write an expression for the cross-sectional area $A (z)$ as a function of z, so that $dv= d^3x=A(z) dz$? All you need then is to proceed with the integral. $\\$ Additional comment: I believe this post might belong in the homework section. If it is homework/schoolwork, please use the Homework section with the homework template on future posts. Thanks. :)

 

[fresh_42]

Additional comment: I believe this post might belong in the homework section. If it is homework/schoolwork, please use the Homework section with the homework template on future posts. Thanks. :)

Right.