Examples of Surfaces

Level Surfaces

A surface can be given as a level surface as

\[\{(x,y,z): f(x,y,z) = 0\]

where \(f\) is smooth.

Claim If \(\Sigma\) is a level surface with smooth \(f\) and \(\forall p \in \Sigma. \nabla f\neq \vec 0\). Then \(\Sigma\) is a regular surface.

proof. Near each point \(\mathbf p=(x_0, y_0, z_0)\in \Sigma, f(\mathbf p) = 0\), we want to show that \(\Sigma\) is a graph of coordinate function of the other two coordinates.
WLOG assuming that \(\sigma\) is the map from XZ-plane.

\[\sigma(x, z) := (x, \psi(x, z), z)\]

where \(\psi(x_0, z_0) = y_0\).

By implicit funciton theorem, consider \(f\) near \(\mathbf p\).

\[\begin{align*} f(\mathbf a) &= f(\mathbf p) + Df\vert_{\mathbf p} (\mathbf a-\mathbf p) + rem.\\ &= 0 + \begin{bmatrix} \frac{\partial f}{\partial x}(x_0, y_0, z_0)\\ \frac{\partial f}{\partial y}(x_0, y_0, z_0)\\ \frac{\partial f}{\partial z}(x_0, y_0, z_0)\end{bmatrix}\begin{bmatrix}x-x_0\\y-y_0\\z-z_0\end{bmatrix} + rem. \end{align*}\]

Note that \(Df\vert_{\mathbf p}\neq 0\) so that the linear expansion is invertible. Therefore, we can apply implicit function theorem and confirm that the non-linear \(f\) is also invertible.

In this case, \(Df|_{\mathbf p}\) is non-zero and is the normal vector of \(\Sigma\). In this case, we also have that \(\Sigma\) is orientable, since \(N(p) = Df|_{\mathbf p}\neq 0\) and \(f\) is smooth.

Examples of level surface

For level surface

\[\Sigma = \{(x,y,z): x^2 + y^2 + z^4 = 1\}\]

we have that \(Df = (2x, 2y, 4z^3)\) vanishes only when at \(p=(0, 0, 0)\), while \(p\not\in\Sigma\), hence it is smooth.

Consider \(a>b>0\) and

\[\Sigma = \{(x,y,z): (x^2+y^2+z^2 + a^2 - b^2)^2 = (4a^2(x^2+y^2))\}\]

Let \(f(x,y,z) = (x^2+y^2+z^2 + a^2 - b^2)^2 - (4a^2(x^2+y^2))\), then

\[Df = \begin{pmatrix} 4x(x^2+y^2+z^2 - a^2 - b^2)\\ 4y(x^2+y^2+z^2 - a^2 - b^2)\\ 4z(x^2+y^2+z^2 + a^2 - b^2) \end{pmatrix}\]

The only possible point where \(Df=0\) is \(p(0,0,0)\not\in\Sigma\), hence \(\Sigma\) is smooth. Furthere more, consider

\[\begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} (a+b\cos\theta)\cos\varphi\\ (a+b\cos\theta)\sin\varphi\\ b\sin\theta \end{pmatrix}\]

Note that

\[x^2+y^2 + z^2 = a^2+2ab\cos\theta + b^2\cos^2\theta + b^2\sin^2\theta = a^2 + 2ab\cos\theta + b^2\]

\[\begin{align*} LHS &= (a^2 + 2ab\cos\theta + b^2 +a^2 - b^2)^2\\ &= (2a)^2(a + b\cos\theta)^2\\ &= 4a^2(x^2 + y^2)\\ &= RHS \end{align*}\]

Lagrange's Method of Undetermined Multipliers

Let \(\Sigma = \{(x,y,z): f(x,y,z) = 0\}\) be smooth.

Claim 1 \(\nabla f\) is perpendicular to the tangent plane at every \(p\in \Sigma\).

proof. Since \(\Sigma\) is smooth. Let \(p\in\Sigma\), let \(\sigma: U\rightarrow V\) be some patch containing \(p\). Then, we have that

\[\frac{df}{du} = \frac{df}{dx}\frac{dx}{du}+\frac{df}{dy}\frac{dy}{du}+\frac{df}{dz}\frac{dz}{du} = \nabla f\cdot \sigma_u\]

\[\frac{df}{dv} = \frac{df}{dx}\frac{dx}{dv}+\frac{df}{dy}\frac{dy}{dv}+\frac{df}{dz}\frac{dz}{dv} = \nabla f\cdot \sigma_v\]

However, \(\Sigma\) is defined on a level surface so that \(\frac{df}{du} = \frac{df}{dv} = 0\). while \(\nabla f\neq 0, D\sigma \neq 0\), hence \(\nabla f\perp \sigma_v\).

Claim 2. \(\Sigma\) is orientable.

proof. Note that \(\nabla f\) is perpendicular to the tangent plane at all \(p \in S\). Which means it is parallel to \(\hat N\). Therefore, we can take \(\hat N = \frac{\nabla f}{\|\nabla f\|}\), since \(f\) is smooth and nowhere vanishing, \(\hat N\) is well-defined and \(\Sigma\) is therefore orientable.

Claim 3. For some \(F:\mathbb R^3\rightarrow \mathbb R\) is smooth, and \(F\), with the restriction to \(\Sigma\), has a local extremum at \(p\), then \(\nabla F = \lambda \nabla f\).

proof. Let \(\gamma\) be arbitrary smooth curve on \(\Sigma\) passing through \(p\), since \(F\) has a local extremum on \(\Sigma\), it is also a extremum on such path \(\gamma\), therefore we have that

\[\frac{dF}{dt} = \nabla f\cdot \frac{d\gamma}{dt} = 0\]

Therefore, \(\nabla F\perp \gamma'\) is perpendicular to any curve on \(\Sigma\) at \(p\), implying that \(\nabla F\) is perpendicular to \(T_p\Sigma\). Therefore \(\nabla F\) parallel to \(\nabla f\implies \nabla f = \lambda \nabla F\)

Surface of Revolution

A surface of revolution is the surface obtained by rotating a plane curve around a straight line in the plane.

Consider the example where the curve resides in the XZ plane, defined as

\[\gamma: (a, b)\rightarrow \mathbb R^3. \gamma(t)= (f(t), 0, g(t)), f(t) > 0\]

and the surface is obtained by rotating \(\gamma\) about z-axis with rotation parameter \(\theta\), the surface patch is then

\[\sigma(t, \theta) = (f(t)\cos\theta, f(t)\sin\theta, g(t))\]

Where \(\sigma_1: U_1\rightarrow V_1\cap \Sigma\) with \(\theta\in (0, 2\pi), \sigma_2: U_2\rightarrow V_2\cap \Sigma\) with \(\theta\in (-\pi, \pi)\)

For example, the sphere without the north and south pole is a surface of revolution with \(\gamma(t) = (\cos t, 0, \sin t), t\in (-\pi/2, \pi/2)\). Cylinder with \(\gamma(t) = (R, 0, t), t\in\mathbb R\)

Note that

\[\begin{align*} \partial_t\sigma &= (f'(t)\cos\theta, f'(t)\sin\theta, g'(t))\\ \partial_\theta\sigma &= (-f(t)\sin\theta, f(t)\cos\theta, 0)\\ N = \sigma_t\times \sigma_\theta &= (f(t)g'(t)\cos\theta, -f(t)g'(t)\sin\theta, f(t)f'(t))\\ \|N\| &= \sqrt{f^2(t)(g'(t))^2 + f^2(t)(f'(t))^2}\\ &= f(t)\sqrt{(f'(t)^2) + (g'(t)^2) }\\ &= f(t)\|\gamma'(t)\| \end{align*}\]

Frequently, \(\gamma\) is parameterized by the arc-length \(s=t\) so that \((f'(t)^2) + (g'(t)^2) = 1\), if so

\[\|N(t,\theta)\| = f(t)\]

Ruled Surfaces

A ruled surface is a union of straight lines called rullings of the surface along some curve \(\gamma\), where the curve is often unit-speed.

Let \(\gamma:(a,b)\rightarrow \mathbb R^3\) parameterizes the curve, and \(d:(a,b)\rightarrow \mathbb R^3\) parameterize the direction of the line. Conveniently, assume \(\gamma, d\) are both unit-speed. Then, the surface is parameterized by

\[\sigma(u, v) = \gamma(u) + vd(u)\]

where \(u\in (a, b), v\in\mathbb R\)

\[\begin{align*} \sigma_u &= \gamma'(u) + vd'(u)\\ \sigma_v &= d(u) \end{align*}\]

Therefore, we need that \(d\neq 0, \gamma'(u) + vd'(u)\) are \(d(u)\) are linearly independent to satisfy the regularity conditions. One way to ensure that is to require \(\gamma'\) and \(d\) being linearly independent and \(v\) sufficiently small, say \(v\in (-\epsilon, \epsilon)\).

Oppositely, one can consider surfaces where \(d\) is parallel to \(\gamma'\) and \(d'\neq 0\) (since \(d\) is unit-speed, this means \(d\perp d'\)).
Then, we can replace \(d\) with \(\gamma'\) so that

\[\sigma_u = \gamma' + vd', \sigma_v = \gamma'\]

is linearly independent IFF \(\gamma'\) is not parallel to \(d'\) IFF \(d\) is not parallel to \(d'\) and \(v\in\mathbb R - \{0\}\)

Examples of Ruled Surface

Mobius band is a ruled surface with

\[\gamma(u) = (\cos u, \sin u, 0), d(v) = (\sin u, 0, \cos u)\]

A generalized cylinder is when the rulings is constant, \(d(u) = \mathbf c\), so that the rulings are always parallel. Then, \(\sigma_u = \gamma'(u), \sigma_v = \mathbf c\) so that \(\sigma\) is regular IFF \(\gamma\) is never tangent to \(\mathbf c\).

A generalized cone with vertex \(\mathbf p\) is when all rulings pass through some fixed point \(\mathbf p\). In this case, \(d(u) = \gamma(u)-\mathbf p\) so that

\[\sigma_u = \gamma'(u) + v\gamma'(u), \sigma_v = \gamma(u)-\mathbf p\]

Therefore, \(\sigma\) is regular IFF \(v\neq -1\), i.e. \(\mathbf p\) is not in the surface.