Verifying Divergence Theorem with Triple/Surface Integrals

[blank_slate]

I am trying to verify the divergence theorem by using the triple integral and the surface integral of the vector field dotted with dS.

No trouble per se, I'm not sure though about one thing: I am given a function and six planes (they form a cube). When I set x=0 the vector field is given as <3,y,2z> and I need to dot that with the normal vector. I am choosing the normal vector as <0,0,0>. I get the same answer as the book I am using, but they chose a different normal vector. Is my normal vector <0,0,0> right when I have a plane x=0?

I am doing the same for other planes: y=0 normal: <0,0,0>; y=1 normal: <0,1,0>; z=1 normal: <0,0,1> etc.

Thanks!

 

[HallsofIvy]

I am trying to verify the divergence theorem by using the triple integral and the surface integral of the vector field dotted with dS.

No trouble per se, I'm not sure though about one thing: I am given a function and six planes (they form a cube). When I set x=0 the vector field is given as <3,y,2z> and I need to dot that with the normal vector. I am choosing the normal vector as <0,0,0>. I get the same answer as the book I am using, but they chose a different normal vector. Is my normal vector <0,0,0> right when I have a plane x=0?

< 0, 0, 0> is not a "unit" normal vector (I started to say not a normal vector but in a sense, it is normal to everything!). If you want an "outward" pointing unit normal vector to the yz-plane (x= 0), then you want <-1, 0, 0>.

I am doing the same for other planes: y=0 normal: <0,0,0>; y=1 normal: <0,1,0>; z=1 normal: <0,0,1> etc.

Thanks!

Those others are also wrong. The outward pointing unit normal to the y= 0 plane is <0, -1, 0>, to z= 0, <0, 0, -1>. I suspect that the only reason you got the right answer was that the function to be integrated happened to be 0 on x= 0, y= 0, and z= 0.