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

question 74

Multiple Choice

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.

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