Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$ and Sis the region bounded by and $x = 0 , x = 3$ . Using the Divergence Theorem, $\iint S$$y = 0 , y = 3 \text {, }$ is

A)203
B)211
C)227
D)243
E)265

Question

Let

\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }

and Sis the region bounded by and

x = 0 , x = 3

. Using the Divergence Theorem,

\iint S

y = 0 , y = 3 \text {, }

is

A)203
B)211
C)227
D)243
E)265

Quizplus · Accepted Answer

The Divergence Theorem relates the flux of a vector field through a closed surface to the divergence of the field inside the volume enclosed by the surface. The divergence of $\mathbf{F}$ is $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}x^2 + \frac{\partial}{\partial y}y^2 + \frac{\partial}{\partial z}z^2 = 2x + 2y + 2z$. The volume integral of the divergence over the region bounded by $x=0, x=3, y=0, y=3, z=0, z=3$ is $\int_0^3 \int_0^3 \int_0^3 (2x + 2y + 2z) \, dz \, dy \, dx$. Solving this gives $243$, matching option D.

Let F(x,y,z)=x2i+y2j+z2k\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }F(x,y,z)=x2i+y2j+z2k

Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$