Solution

A subset

S \subset R^{3}

is a smooth surface if every point has a neighbourhood in

S

parametrized by a map

X : U \to S

, where

U \subset R^{2}

is open,

X

is a homeomorphism onto that neighbourhood,

X

is smooth, and

D X

has rank two everywhere. These are the embedded surface parametrization conditions.

For the given set use local angular intervals in the parametrization

X (θ, t) = (ϕ (t) cos θ, ϕ (t) sin θ, t) .

It is locally one-to-one and has tangent vectors

X_{θ} = (- ϕ sin θ, ϕ cos θ, 0), X_{t} = (ϕ^{'} cos θ, ϕ^{'} sin θ, 1) .

Their cross product has magnitude

∣ X_{θ} \times X_{t} ∣ = ϕ (t) 1 + ϕ^{'} (t)^{2} > 0,

so the derivative has rank two. The angular charts cover

Σ

, proving that it is a smooth surface.

The area between heights

a_{0}

and

b_{0}

is therefore

2 π \int_{a_{0}}^{b_{0}} ϕ (t) 1 + ϕ^{'} (t)^{2} d t .

By hypothesis this equals

2 π r (b_{0} - a_{0})

for every subinterval. Since the integrand is continuous,

ϕ (t) 1 + ϕ^{'} (t)^{2} = r,

and squaring gives

r^{2} = ϕ (t)^{2} + ϕ (t)^{2} ϕ^{'} (t)^{2} .

0 < ϕ (t) < r

, then

ϕ^{'}

never vanishes, so its sign

σ \in {1, - 1}

is constant. The last equation gives

ϕ^{'} = σ \frac{r ^{2} - ϕ ^{2}}{ϕ},

and hence

\frac{d}{d t} r^{2} - ϕ (t)^{2} = - σ .

Thus

r^{2} - ϕ (t)^{2} = - σ (t - t_{0})

for a constant

t_{0}

, and

(t - t_{0})^{2} + ϕ (t)^{2} = r^{2} .

The graph lies on a circle of radius

r

Solved by gpt-5.6-sol high.

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