Using the Divergence Theorem, find the outward flux of F across the boundary of the region D.
-$$\mathbf { F } = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z \mathbf { k } \text {; D: the solid cube cut by the coordinate planes and the planes } x = 3 , y = 3 \text {, and } z = 3$$
A) 189
B) 164
C) 81
D) 108

Question

Quizplus · Accepted Answer

The Divergence Theorem states that the outward flux of a vector field $\mathbf{F}$ across the boundary of a region $D$ is equal to the triple integral over $D$ of the divergence of $\mathbf{F}$. The divergence of $\mathbf{F} = x^2\mathbf{i} + y^2\mathbf{j} + z\mathbf{k}$ is $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z) = 2x + 2y + 1$. Integrating this over the cube from 0 to 3 for each variable gives the total flux: $\int_0^3 \int_0^3 \int_0^3 (2x + 2y + 1) dzdydx = \int_0^3 \int_0^3 [3(2x + 2y + 1)] dydx = \int_0^3 [9(2x + 3) + 9] dx = 18\int_0^3 x dx + 54\int_0^3 dx = 9(3^2) + 54(3) = 81 + 108 = 189$.

Using the Divergence Theorem, Find the Outward Flux of F