Flux integral

Jun 2012
1
0
The Netherlands
I have the following problem:

Find the flux of F = xi + yj + zk out of the tetrahedron bounded by the coordinate planes and the plane x + 2y + 3z = 6

I got the right answer with the Divergence Theorem, but I have to be able to do it with "direct" calculation of the flux integral. I tried it in different ways but never got to the right answer. Anyone good at this? :)
 

Imo

Jun 2012
7
0
Croatia
i would go like this:
F*dS=∫∫F*(∂r/∂u x ∂r/∂v)dudv
with parametrization:
x=u [0<u<6], y=v [0<v<3-u/2], z=2-u/3-2v/3
r(u,v)=ui+vj+(2-u/3-2v/3)k
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Much the same as Imo said. The tetrahedron has four faces and to find the flux directly you will need to integrate over all four faces.
Three of the faces are in the coordinate planes. Of course, the plane x+ 2y+ 3z= 6 crosses the three axes at (6, 0, 0), (0, 3, 0) and (0, 0, 2) so the face in the xy plane (z= 0) is the triangle with vertices (0, 0, 0), (6, 0, 0), and (0, 3, 0). \(\displaystyle d\vec{S}= <0, 0, 1>dxdy\) so you need to integrate <x, y, z>.<0, 0, 1>dxdy= zdxdy= 0 (because z= 0 in this plane) over that triangle which just gives 0. Similarly for the other two sides in the coordinate planes.

That is why you really only need to integrate on the x+ 2y+ 3z= 6 face. That can be written x= 6- 2y- 3z or, as a vector formula, \(\displaystyle \vec{r}(y,z)= (6- 2y- 3z)\vec{i}+ y\vec{j}+ z\vec{k}\). The derivative with resepect to y is \(\displaystyle -2\vec{i}+ \vec{j}\) and with respect to z, \(\displaystyle -3\vec{i}+ \vec{k}\), so the "\(\displaystyle \partial r/\partial u\times \partial r/\partial v\)" of Imo is
\(\displaystyle \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k}\\ -2 & 1 & 0 \\ -3 & 0 & 1\end{array}\right|= \vec{i}+ 2\vec{j}+ 3\vec{k}\)
Integrate \(\displaystyle (x\vec{i}+ y\vec{j}+ z\vec{k})\cdot(\vec{i}+ 2\vec{j}+ 3\vec{k})dydz\) over the triangle, in the yz-plane, with vertices at (0, 0, 0), (0, 3, 0), and (0, 0, 2).
 
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