- #1

BloonAinte

- 19

- 1

- Homework Statement
- Prove ##\int F \times dr## = ##\alpha \int (\nabla \times F) \times dS##

where ##\alpha## is a constant to be found. Here $F$ is a vector field, S is an open surface and C is the boundary of S.

- Relevant Equations
- Stokes' theorem: ##\int_C F \cdot dr = \int_S \nabla \times F \cdot dS##

I considered the vector field ##a \times F##, and applied Stoke's theorem. I obtained that $$\int_C (a \times F) \cdot dr = \int_C (F \times dr ) \cdot a.$$

Now, $$\nabla \times (a \times F) = a (\nabla \cdot F) - (a \cdot \nabla) F.$$

Using Stoke's theorem for the vector field ##a \times F##, I obtain $$\int_C (F \times dr ) \cdot a = \int_C (a \times F) \cdot dr = \int_S \nabla \times (a \times F) \cdot dS = \int_S [a (\nabla \cdot F) - (a \cdot \nabla) F] \cdot dS$$

But then, I cannot proceed. Can someone give me a hint? I want to get the surface integral into the form ##a \cdot \int_S (\nabla \times F) \times dS##

Now, $$\nabla \times (a \times F) = a (\nabla \cdot F) - (a \cdot \nabla) F.$$

Using Stoke's theorem for the vector field ##a \times F##, I obtain $$\int_C (F \times dr ) \cdot a = \int_C (a \times F) \cdot dr = \int_S \nabla \times (a \times F) \cdot dS = \int_S [a (\nabla \cdot F) - (a \cdot \nabla) F] \cdot dS$$

But then, I cannot proceed. Can someone give me a hint? I want to get the surface integral into the form ##a \cdot \int_S (\nabla \times F) \times dS##

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