below math question Evaluate the surface integralF• ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use thepositive (outward) orientation.xze' i - xze j + z kF(x, y, z)S is the part of the plane x + y + z = 5 in the first octant and has downward orientation

It is given that for a vector field F(x,y,z)=xzeyi-xzeyj+zk and oriented surface S which is part of the plane x+y+z=5 lies in first octant. The objective is to evaluate the surface integral ∬SF.dS. The surface integral of a function F over a surface S which is defined as z=gx,y is expressed as, ∬SF.dS=∫∫RFx,y,g(x,y)gxx,y2+gyx,y2+1dASince, S is part of plane x+y+z=5 in first octant. Therefore, z=gx,y=5-x-y We have, gx=-1gy=-1 So, that dS=gx2+gy2+1dA=1+1+1dA=3dA The surface integral is, ∬SF.dS=∫∫Rx5-x-yey-x5-x-yey+5-x-y3dA where, z=5-x-y For the limits of integration, When z=0, We have, x+y=5 So, y varies from 0 to 5-x and plane lies in first octant according to question. And x from 0 to 5. Simplify the integral as, ∬SF.dS=∫∫Rx5-x-yey-x5-x-yey+5-x-y3dA=∫05∫05-x5-x-y3dydx=3∫055y-xy-y2205-xdx=3∫0555-x-x5-x-5-x22dx On further simplification, we have ∬SF.dS=3∫05252-5x+x22dx=325x2-5x22+x3605=32552-5522+536-0=12536 Answer: The value of surface integral is ∬SF.dS=12536.