Let S be the part of the plane 2x+5y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field 1i+3j+1k across the surface S.

Answer 1

You have to compute ∫∫_SF·dS. We can parameterize the surface given by r(u,v)=<u,v,4-2u-5v> for 0<u<2 and 0<v<-2/5u+4/5. We have r_u(u,v)=<1,0,-2> and r_v(u,v)=<0,1,-5>, so that r_u(u,v) x r_v(u,v)=<2,5,1> •Therefore

∫∫_SF·dS=∫∫_SF•(r_u(u,v) x r_v(u,v))dA=∫∫_S<1,3,1>•<2,5,1>dA=∫∫_S18dA=18 ∫_{0}^{2} ∫_{0}^{-2/5u+4/5}dvdu. You should be able ti compute this double integral!