Evaluate the surface integral S F · 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 the positive (outward) orientation. F(x, y, z) = x i + y j + 7 k S is the boundary of the region enclosed by the cylinder x2 + z2 = 1 and the planes y = 0 and x + y = 6
V is the 3D region enclosed by S.
div(F)= ∂(x)/∂x + ∂(y)/∂y + ∂(10)/∂z = 2
∫ = ∫∫∫ [V] div(F) dV = 2 ∫∫∫ [V] dV
Using cylindrical coords, ( rcos(θ), y, rsin(θ) ), dV=rdrdθdy
∫ = 2 ∫∫∫ rdrdθdy, [r=0,1], [θ=0,2π], [y=0,7−rcos(θ)] = 14π (easily integrated)
Answer:
z=(x^2+y^2)
given equation
x^2+z^2=1
and y=0 and
x+y=6
For a given problem for the given vector calculus,the final value of integration is 45.56