Find the outward flux of F(x, y, z) = 5(x

Question

Find the outward flux of F(x, y, z) = 5(x - 1)i + 5(y - 3)j + 5z k across the rectangular box with vertices (0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (4, 0, 0), (4, 1, 0), (4, 0, 1), and (4, 1, 1).

Quizplus · Accepted Answer

By using the Divergence Theorem, we can calculate the outward flux as the triple integral of the divergence of F over the region enclosed by the rectangular box:

∬∬F⋅n dS = ∭div(F) dV

where n is the unit outward normal vector to the surface element dS.

The divergence of F is simply 5 + 5 + 5 = 15, since div(F) = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z.

The region enclosed by the rectangular box is a parallelepiped with six faces, each of which is a rectangle parallel to one of the coordinate planes. Hence, we can split the triple integral into six separate integrals, one for each face:

∭div(F) dV = ∫0⁴ ∫0¹ ∫0¹ (5 + 0 + 0) dy dz dx    (integral over the left face)
                      + ∫0⁴ ∫0¹ ∫0¹ (0 + 5 + 0) dy dz dx    (integral over the back face)
                      + ∫0⁴ ∫0¹ ∫0¹ (-5 + 0 + 0) dy dz dx    (integral over the right face)
                      + ∫0⁴ ∫0¹ ∫0¹ (0 - 5 + 0) dy dz dx    (integral over the front face)
                      + ∫0¹ ∫0⁴ ∫0¹ (0 + 0 + 0) dx dz dy    (integral over the bottom face)
                      + ∫0¹ ∫0⁴ ∫0¹ (0 + 0 + 5z) dx dz dy    (integral over the top face)
                      = 6(5)(4)(1)          (using the linearity of the integral)
                      = 120

Therefore, the outward flux of F across the rectangular box is 120, which is answer choice B.