13.7no2

Can you help me with Let S be the part of the plane 3x+2y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=1i+1j+4k across the surface S.

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OK, here's an answer in general terms, so I don't do your homework problem for you.  You can write the plane as the graph z=f(x,y) of the function f(x,y)=4-3x-2y.  You can then use the usual parameterization of the graph of a function: r(u,v)=<u,v,4-3u-4v>, and then compute that
  r_u x r_v = ai+bj+ck, 
for whatever numbers a,b,c you get from computing the cross product.  Since the orientation is upward, you have to be a little careful: c has to be positive, so if   r_u x r_v  has c negative you have to use the -(r_u x r_v) instead to get the upward orientation.  The shadow of S in the xy-plane is the triangle enclosed by the x-axis, the y-axis, and the line 4-3x-2y=0, and the same triangle is the domain D of the parameterization. Then ∫∫_S F.dS=∫∫_D ∫∫_S F.(r_u x r_v) dA