Discussion
The flux of \( \mathbf{F} \) through a surface \( S \) is
Answer: C
Only the component crossing the surface counts, so each element contributes \( \mathbf{F} \cdot \mathbf{n}\,dS \) and
\[ \Phi \;=\; \iint_{S} \mathbf{F} \cdot \mathbf{n}\,dS. \]
A field lying wholly in the surface carries no flux through it, which the dot product records and an integral of the magnitude would not
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