Paper 3, Section II, B

Vector Calculus | Part IA, 2017

By a suitable choice of u\mathbf{u} in the divergence theorem

VudV=SudS\int_{V} \nabla \cdot \mathbf{u} d V=\int_{S} \mathbf{u} \cdot d \mathbf{S}

show that

VϕdV=SϕdS\int_{V} \nabla \phi d V=\int_{S} \phi d \mathbf{S}

for any continuously differentiable function ϕ\phi.

For the curved surface of the cone

x=(rcosθ,rsinθ,3r),03r1,0θ2π\mathbf{x}=(r \cos \theta, r \sin \theta, \sqrt{3} r), \quad 0 \leqslant \sqrt{3} r \leqslant 1, \quad 0 \leqslant \theta \leqslant 2 \pi

show that dS=(3cosθ,3sinθ,1)rdrdθd \mathbf{S}=(\sqrt{3} \cos \theta, \sqrt{3} \sin \theta,-1) r d r d \theta.

Verify that ()(*) holds for this cone and ϕ(x,y,z)=z2\phi(x, y, z)=z^{2}.

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