Let $\vec { a } = \alpha _ { 1 } \vec { i } + \alpha _ { 2 } \vec { j } + \alpha _ { 3 } \vec { k }$ be a nonzero constant vector and let $\vec { r } = x \vec { i } + y \vec { j } + z \vec { k }$ .Suppose S is the sphere of radius one centered at the origin.There are two (related)reasons why $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ .Select them both.
A) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\vec { a } \times \vec { r } = \overrightarrow { 0 }$ .
B) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\operatorname { div } ( \vec { a } \times \vec { r } ) = 0$ .
C) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\vec { r }$ is parallel to the
$d \vec { A }$ element everywhere on S and so
$\vec { a } \times \vec { r }$ is perpendicular to
$d \vec { A }$ on S.
D) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\vec { a } \times \vec { r }$ is a constant vector field.
E) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\int _ { H } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = Q 0$ and
$\int _ { L } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = - Q$ , where H is the upper unit hemisphere and L is the lower unit hemisphere.

Question

Let $\vec { a } = \alpha _ { 1 } \vec { i } + \alpha _ { 2 } \vec { j } + \alpha _ { 3 } \vec { k }$ be a nonzero constant vector and let $\vec { r } = x \vec { i } + y \vec { j } + z \vec { k }$ .Suppose S is the sphere of radius one centered at the origin.There are two (related)reasons why $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ .Select them both.
A) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\vec { a } \times \vec { r } = \overrightarrow { 0 }$ .
B) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\operatorname { div } ( \vec { a } \times \vec { r } ) = 0$ .
C) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\vec { r }$ is parallel to the
$d \vec { A }$ element everywhere on S and so
$\vec { a } \times \vec { r }$ is perpendicular to
$d \vec { A }$ on S.
D) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\vec { a } \times \vec { r }$ is a constant vector field.
E) $\int _ { S } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = 0$ because $\int _ { H } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = Q > 0$ and
$\int _ { L } ( \vec { a } \times \vec { r } ) \cdot d \vec { A } = - Q$ , where H is the upper unit hemisphere and L is the lower unit hemisphere.

Quizplus · Accepted Answer

The divergence theorem says that the flux of a vector field through a closed surface is equal to the triple integral of the divergence of the vector field over the region enclosed by the surface. In this case, the divergence of $\vec { a } \times \vec { r }$ is zero, so the flux of $\vec { a } \times \vec { r }$ through the sphere is zero.Another way to see this is that $\vec { a } \times \vec { r }$ is perpendicular to $\vec { r }$ everywhere on the sphere. This means that the dot product of $\vec { a } \times \vec { r }$ with the outward unit normal vector to the sphere is zero everywhere on the sphere. Therefore, the flux of $\vec { a } \times \vec { r }$ through the sphere is zero.

Let a⃗=α1i⃗+α2j⃗+α3k⃗\vec { a } = \alpha _ { 1 } \vec { i } + \alpha _ { 2 } \vec { j } + \alpha _ { 3 } \vec { k }a=α1​i+α2​j​+α3​k

Let $\vec { a } = \alpha _ { 1 } \vec { i } + \alpha _ { 2 } \vec { j } + \alpha _ { 3 } \vec { k }$