Find the flux of the curl of field F through the shell S.
-$\mathbf { F } = - 7 \mathrm { zi } + 6 x \mathbf { j } + 4 y \mathbf { k } ; S$ is the portion of the cone $z = 6 \sqrt { x ^ { 2 } + y ^ { 2 } }$ below the plane $z = 2$
A) $- \frac { 2 } { 3 } \pi$
B) $\frac { 4 } { 3 } \pi$
C) $- \frac { 1 } { 3 } \pi$
D) $- \frac { 4 } { 3 } \pi$

Question

Quizplus · Accepted Answer

The curl of F is $\mathbf { curl } \mathbf { F } = - 6 \mathbf { j } + 4 \mathbf { k }$. The surface S is oriented upward, so the normal vector is $\mathbf { n } = \frac { 1 } { \sqrt { 36 + 1 } } ( - 6 \mathbf { i } + 4 \mathbf { j } + 6 \mathbf { k } )$. Therefore, the flux of the curl of F through S is$$\begin{split} \iint\limits_S (\mathbf { curl } \mathbf { F } ) \cdot \mathbf { n } d \sigma & = \iint\limits_S (- 6 \mathbf { j } + 4 \mathbf { k } ) \cdot \frac { 1 } { \sqrt { 36 + 1 } } ( - 6 \mathbf { i } + 4 \mathbf { j } + 6 \mathbf { k } ) d \sigma \\ & = \iint\limits_S \frac { 1 } { \sqrt { 36 + 1 } } ( 36 - 16 ) d \sigma \\ & = \frac { 20 } { \sqrt { 36 + 1 } } \iint\limits_S d \sigma \\ & = \frac { 20 } { \sqrt { 36 + 1 } } \cdot \pi ( 6 \sqrt { 2 } )^2 \\ & = - \frac { 2 } { 3 } \pi. \end{split}$$

Find the Flux of the Curl of Field F Through F=−7zi+6xj+4yk;S\mathbf { F } = - 7 \mathrm { zi } + 6 x \mathbf { j } + 4 y \mathbf { k } ; SF=−7zi+6xj+4yk;S

Find the Flux of the Curl of Field F Through $\mathbf { F } = - 7 \mathrm { zi } + 6 x \mathbf { j } + 4 y \mathbf { k } ; S$