063ccd…: “Such as the funny town”

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In the exercises that follow, consider the function $\vec {F} : \R ^3 \to \R ^3$ defined by the rule $\vec {F}(x,y,z) = \vector {z,x + y + z,y + 2 z}$ .

Compute $\divergence \vec {F}$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

$\divergence \vec {F} = \answer { 3 }$

Compute $\curl \vec {F}$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

$\curl \vec {F} = \vector {\answer { 0 },\answer { 1 },\answer { 1 }}$

Compute $\grad \vec {F}$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

Compute $\curl \left ( \grad \vec {F}\right )$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

Compute $\divergence \left ( \grad \vec {F}\right )$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

Compute $\grad \left ( \divergence \vec {F}\right )$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

Is the case for any vector field $\vec {F} : \R ^3 \to \R ^3$ that $\grad \left ( \divergence \vec {F}\right ) = \vec {0}$ ?

No. Yes.

Compute $\curl \left ( \curl \vec {F}\right )$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

Is the case for any vector field $\vec {F} : \R ^3 \to \R ^3$ that $\curl \left ( \curl \vec {F}\right ) = \vec {0}$ ?

No. Yes.

Compute $\divergence \left ( \curl \vec {F}\right )$ .

This is zero. This can be computed (and is nonzero). That does not even make sense.

Is the case for any vector field $\vec {F} : \R ^3 \to \R ^3$ that $\divergence \left ( \curl \vec {F}\right ) = 0$ ?

No. Yes.