Chavez Bryant
06/26/2024 · High School
prove: \[ \int_{T} \nabla \phi d T=\int_{\partial T} \phi d \partial \text { and } \int_{T} \nabla \times V d T=-\int_{\partial T} v \times d \sigma \] [Hint: Apply Gauss's Theorem \( \int_{T} \nabla \cdot F d T=\int_{\partial T} F \cdot d \sigma \) to \( F=\phi e_{x} \) and \( F=v \times e_{x} \), etc.]
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The proof involves applying Gauss's Theorem to the vector fields \( F = \phi \mathbf{e}_x \) and \( F = V \times \mathbf{e}_x \). For the first statement, the divergence of \( F \) is computed and related to the surface integral of \( F \). For the second statement, the divergence of the cross product is used to relate the volume integral to the surface integral. The results show that:
1. \(\int_{T} \nabla \phi \, dT = \int_{\partial T} \phi \, d\sigma\)
2. \(\int_{T} \nabla \times V \, dT = -\int_{\partial T} v \times d\sigma\)
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