Final answer:
To compute the surface integral ∫∫ⁿⁿ F · dS, where F(x, y, z) = xyi + zj + (x + y)k and N is the outward normal vector to the triangular region S, cut off from the plane x + y + z = 1 by the positive coordinate axes, you can use the divergence theorem. The surface integral can be evaluated by converting it into a triple integral over the region enclosed by the surface. The triple integral is computed by integrating over the three coordinate axes.
Step-by-step explanation:
To compute the surface integral ∫∫ⁿⁿ F · dS, where F(x, y, z) = xyi + zj + (x + y)k and N is the outward normal vector to the triangular region S, cut off from the plane x + y + z = 1 by the positive coordinate axes, we can use the divergence theorem. The divergence theorem states that the surface integral of a vector field F over a closed surface S is equal to the triple integral of the divergence of F over the region enclosed by the surface. In this case, the region enclosed by the surface S is a tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1).
- Using the divergence theorem, the triple integral of the divergence of F = (d/dx)(xy) + (d/dy)(z) + (d/dz)(x + y) over the tetrahedron is equal to the surface integral of F · dS over the triangular region S. The divergence of F is equal to y + 1. So, the triple integral becomes ∭ (y + 1) dV, where dV represents an infinitesimal volume element.
- Evaluating the integral ∭ (y + 1) dV over the tetrahedron is equivalent to integrating over the three coordinate axes. This can be done by integrating with respect to y first, then x, and finally z. The limits of integration for each variable are determined by the constraints of the tetrahedron.
- Computing the triple integral ∭ (y + 1) dV over the tetrahedron yields the value of the surface integral ∫ ∫ⁿⁿ F · dS over the triangular region S. This value represents the flux of the vector field F through the surface S.