Let $\mathbf { F } ( x , y , z ) = x \mathbf { i } + y \mathbf { j } + z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ . Using the Divergence Theorem, $\iint S$$\mathbf { F } \bullet \mathbf { n } d S$ is

A) $24 \pi$
B) $28 \pi$
C) $32 \pi$
D) $42 \pi$
E) $48 \pi$

Question

Let

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

and S is

x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4

. Using the Divergence Theorem,

\iint S

\mathbf { F } \bullet \mathbf { n } d S

is

A) $24 \pi$
B) $28 \pi$
C) $32 \pi$
D) $42 \pi$
E) $48 \pi$

Quizplus · Accepted Answer

The divergence of $\mathbf{F}$ is $\nabla \cdot \mathbf{F} = 3$. The volume of the sphere $x^2 + y^2 + z^2 = 4$ is $\frac{4}{3}\pi(2)^3 = \frac{32}{3}\pi$. By the Divergence Theorem, $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = 3 \times \frac{32}{3}\pi = 32\pi$.

Let F(x,y,z)=xi+yj+zk\mathbf { F } ( x , y , z ) = x \mathbf { i } + y \mathbf { j } + z \mathbf { k }F(x,y,z)=xi+yj+zk

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