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 $$ F = V 	imes \mathbf{e}_x $$. 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. $$\int_{T} 
abla \phi \, dT = \int_{\partial T} \phi \, d\sigma$$ 2. $$\int_{T} 
abla 	imes V \, dT = -\int_{\partial T} v 	imes d\sigma$$

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