Choose the correct answer below. B. Yes. The integrand |grad*F| has an upper bound of 1 because |F|

Question

Choose the correct answer below. B. Yes. The integrand

|grad*F|

has an upper bound of 1 because

|F|<=1

. Therefore, the magnitude of the flux has an upper bound equal to the volume of

D

. C. Yes. The flux equals the integral of

F*n

over

S

by the Divergence Theorem.

|F*n|

has an upper bound of 1 because

|F|<=1

and

n

is a unit vector field. Therefore the magnitude of the flux has an upper bound equal to the volume of D . D. Yes. The integrand

|grad*F|

has an upper bound of 1 because

|F|<=1

. Therefore, the magnitude of the flux has an upper bound equal to the surface area of

S

. E. No. Even though

F

has an upper bound

|F|<=1

, the integrand of the volume integral

|grad*F|

is unbounded and the integrand of the surface integral

|F*n|

is unbounded.

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Solution for - Choose the correct answer below. B. Yes. The integrand |grad*F| has an upper bound of 1 because |F|&

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(Solved): Choose the correct answer below. B. Yes. The integrand |grad*F| has an upper bound of 1 because |F|& ...