Let $\mathbf { F } ( x , y , z ) = 2 z \mathbf { i } + 3 x \mathbf { j } + 4 z \mathbf { k }$ and C be the boundary of $z = 4 - x ^ { 2 } - y ^ { 2 }$ above z = 0. Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

A)12?
B)15?
C)18?
D)24?
E)28?

Question

Let

\mathbf { F } ( x , y , z ) = 2 z \mathbf { i } + 3 x \mathbf { j } + 4 z \mathbf { k }

and C be the boundary of

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

above z = 0. Using Stokes' Theorem,

\int _ { C } \mathbf { F } \cdot \mathbf { T } d s

is

A)12?
B)15?
C)18?
D)24?
E)28?

Quizplus · Accepted Answer

Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s = \iint _ { S } ( \nabla \times \mathbf { F } ) \cdot \mathbf { n } d \sigma$, where $S$ is the surface bounded by $C$.$\nabla \times \mathbf { F } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ \frac { \partial } { \partial x } & \frac { \partial } { \partial y } & \frac { \partial } { \partial z } \\ 2 z & 3 x & 4 z \end{array} \right| = -3 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k }$The surface $S$ is given by $z = 4 - x ^ { 2 } - y ^ { 2 }$, so $\mathbf { n } = \frac { - \nabla f } { \Vert \nabla f \Vert } = \frac { - ( - 2 x \mathbf { i } - 2 y \mathbf { j } + \mathbf { k } ) } { \sqrt { 4 x ^ { 2 } + 4 y ^ { 2 } + 1 } } = \frac { 2 x \mathbf { i } + 2 y \mathbf { j } - \mathbf { k } } { \sqrt { 4 x ^ { 2 } + 4 y ^ { 2 } + 1 } }$Therefore, $\iint _ { S } ( \nabla \times \mathbf { F } ) \cdot \mathbf { n } d \sigma = \iint _ { R } ( - 3 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } ) \cdot \frac { 2 x \mathbf { i } + 2 y \mathbf { j } - \mathbf { k } } { \sqrt { 4 x ^ { 2 } + 4 y ^ { 2 } + 1 } } d A$$$= \iint _ { R } \frac { - 6 x + 4 y + 3 } { \sqrt { 4 x ^ { 2 } + 4 y ^ { 2 } + 1 } } d A$$$$= \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 2 } \frac { - 6 r \cos \theta + 4 r \sin \theta + 3 } { \sqrt { 4 r ^ { 2 } + 1 } } r d r d \theta$$$$= \int _ { 0 } ^ { 2 \pi } \left[ - 3 r ^ { 2 } \cos \theta + 2 r ^ { 2 } \sin \theta + 3 r \right] _ { 0 } ^ { 2 } d \theta$$$$= \int _ { 0 } ^ { 2 \pi } ( - 12 \cos \theta + 8 \sin \theta + 6 ) d \theta$$$$= \left[ - 12 \sin \theta - 8 \cos \theta + 6 \theta \right] _ { 0 } ^ { 2 \pi }$$$$= 12$$

Let F(x,y,z)=2zi+3xj+4zk\mathbf { F } ( x , y , z ) = 2 z \mathbf { i } + 3 x \mathbf { j } + 4 z \mathbf { k }F(x,y,z)=2zi+3xj+4zk

Let $\mathbf { F } ( x , y , z ) = 2 z \mathbf { i } + 3 x \mathbf { j } + 4 z \mathbf { k }$