Solution

Expanding with the Levi-Civita symbol and the product rule gives

\nabla \times (a \times b) = a \nabla \cdot b - b \nabla \cdot a + (b \cdot \nabla) a - (a \cdot \nabla) b .

\int_{S} (\nabla \times F) \cdot n d S = \oint_{C} F \cdot d x,

where the boundary orientation follows the right-hand rule about

n

Apply it to

F = m \times v

. On

S

m = n

and

v \cdot n = 0

. Substitution of the vector identity and contraction with

n

leaves

(δ_{ij} - n_{i} n_{j}) \partial_{j} v_{i} .

The boundary integrand satisfies

(n \times v) \cdot d x = v \cdot (d x \times n)

, proving the formula. The vector

d x \times n

is the outward co-normal in the tangent plane of

S

For the hemisphere,

n = e_{r}

and

v = r sin θ e_{θ}

. The projected divergence is

2 cos θ

, so the left side is

\int_{0}^{2 π} \int_{0}^{π /2} 2 cos θ R^{2} sin θ d θ d ϕ = 2 π R^{2} .

On the positively oriented equator,

d x = R e_{ϕ} d ϕ

, hence

d x \times n = R e_{θ} d ϕ

and the boundary integral is also

2 π R^{2}

Solved by gpt-5.6-sol high.

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