(1,3,1)*(2,1,1)/(2*2+1+1)^(1/2)=6/6^(1/2)
Let S be the part of the plane Find the flux of the vector field across the surface S.
I'm stuck
If i use the equation int int Fnds i get the int int of 11/(11)^(1/2)
if I use int int Fn |ruXrv|i get stuck at ruXrv because I don't know how to find my ru and rv.
I can't use the divergence therem because there are no values.
What method should I use? And how do I set up the bounds? Thank you so much!
Sorry for my English is not very well.
The direction of the flux vector is A=(1,3,1),and the direction of the normal of the plane is B=(2,1,1).So the flux of the vector field across the surface need to calculate the flux vector project on the normal of the plane.The angle x formed by two vector A and B is calculated by
cosx=A*B/(׀A׀*׀B׀)
so the angle x between the vector (1,3,1) and (2,1,1) is
cosx=6/66^(1/2)
so the length of the projection on the the plane's normal of the flux vector is
L=׀A׀*cosx=11^(1/2)*6/66^(1/2)=6/6^(1/2)
its direction as same as (2,1,1).