Solution

On the part with

1 \leq z \leq 2

, write

r (ρ, ϕ) = (ρ cos ϕ, ρ sin ϕ, ρ^{2} - 1), 2 \leq ρ \leq 5 .

This is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the

z

d S = (r_{ϕ} \times r_{ρ}) d ρ d ϕ = (\frac{ρ ^{2} cos ϕ}{ρ ^{2} - 1}, \frac{ρ ^{2} sin ϕ}{ρ ^{2} - 1}, - ρ) d ρ d ϕ .

Its radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

For

A = (- y z^{2}, x z^{2}, 0),

the curl is

\nabla \times A = (- 2 x z, - 2 yz, 2 z^{2}) .

On the surface,

z^{2} = ρ^{2} - 1

, and therefore

(\nabla \times A) \cdot d S = - 2 ρ (2 ρ^{2} - 1) d ρ d ϕ .

Direct integration gives

\int_{S} \nabla \times A \cdot d S = - 2 \int_{0}^{2 π} \int_{2}^{5} ρ (2 ρ^{2} - 1) d ρ d ϕ = - 2 π [ρ^{4} - ρ^{2}]_{2}^{5} = - 36 π .

To verify this with the Stokes theorem, note that on a circle of fixed

ρ, z

A = z^{2} ρ e_{ϕ}, d r = ρ e_{ϕ} d ϕ .

The induced boundary orientation runs in the negative

ϕ

direction on the top circle and the positive

ϕ

direction on the bottom circle. Hence

\oint_{\partial S} A \cdot d r = 2 π (1^{2}) (2)^{2} - 2 π (2^{2}) (5)^{2} = 4 π - 40 π = - 36 π .

For

S^{'}

, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive

ϕ

direction at

z = - 1

, where

ρ^{2} = 2

, and negative

ϕ

direction at

z = 2

, where

ρ^{2} = 3

. Thus

\int_{S^{'}} \nabla \times A \cdot d S = 2 π (1) (2) - 2 π (2) (3) = - 8 π .

Solved by gpt-5.6-sol high.

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