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Section 7.2 Area of Parametric Surfaces

For parametric curves, the length of the tangent gave the speed of movement along the curve. For parametric surfacaes, there is no longer a unique sense of movement along the surface, due to the multiple parameters. However, insteed of length, it would be good to have a sense of area in terms of the parametrization. Instead of the tangent vector, the calculus of parametric surfaces is controlled by the normal vector. The length of the normal vector does, indeed, measure area exactly the same way that the length of the tangent vector to a parametric curves measures length.

Let \(\sigma\) be the graph of a differentiable function \(f: \RR^2 \rightarrow \RR\) as defined in Example 7.1.2. Then I can use the area formula to write the general equation of the surface area of the graph of \(f\) over a simply-connected region \(D\) in its domain.

\begin{equation*} A = \int_D \sqrt{ \left( \frac{\del f}{\del x} \right)^2 + \left( \frac{\del f}{\del y} \right)^2 + 1} dx dy \end{equation*}

Let \(f: [a,b] \rightarrow \RR\) be a differentiable function and let \(\sigma\) be the surface of revolution as defined in Example 7.1.3. Let \(D = [a,b] \times [0,2\pi]\text{.}\) Then I can define the surface area of such a general surface of revolution.

\begin{equation*} A = \int_D \sqrt{ 1 - \left( \frac{\del f}{\del x} \right)^2} f(x) dx \end{equation*}

The parabaloid \(z = k(x^2 +y^2)\) can be described by \(\sigma(u,v) = (u,v,k(u^2+v^2))\) over \(D\) a circle of radius \(R\text{.}\) I can calculate its surface area. The height is detemined by the radius of the domain \(D\text{.}\)

Since the domain is a circle, I integrate in polar coordinates. After switching to polar coordinates, a substitution is required so solve the first of the two iterated integrals. In polar coordinates, the two integrals are seperable and the first integral is an integral of just \(d\theta\text{,}\) giving a simple result of \(2\pi\text{.}\)

\begin{align*} \sigma_u \amp = (1, 0, 2ku) \\ \sigma_v \amp = (0, 1, 2kv) \\ \sigma_u \times \sigma_v \amp = (-2ku, -2kv, 1) \\ |\sigma_u \times \sigma_v| \amp = \sqrt{4k^2u^2 + 4k^2v^1 + 1} \\ A \amp = \int_D \sqrt{4k^2 u^2 + 4k^2 v^2 + 1} du dv\\ \amp = \int_0^{2\pi} \int_0^R \sqrt{4k^2 r^2 + 1} r dr d \theta\\ u \amp = 4k^2 r^2 + 1 \implies du = 8k^2 r dr\\ u (R) \amp = 4k^2 R^2 + 1 \\ u (0) \amp = 1 \\ \amp = 2\pi \int_1^{4k^2R^2 + 1} \sqrt{u} \frac{1}{8k^2} du\\ \amp = \left. \frac{\pi}{4k^2} \frac{2u^{\frac{3}{2}}}{3} \right|_1^{4k^2R^2+1} \\ \amp = \frac{\pi}{6k^2} \left( (4k^2R^2 + 1)^{\frac{3}{2}} -1 \right) \end{align*}

The surface of revolution under \(f(x) = \frac{1}{x}\) for \(x \in [1,\infty)\) is called the Horn of Gabriel. It can be described parametrically for \((x,\theta) \in [1, \infty) \times [0, 2\pi]\) as the following surface.

\begin{equation*} \sigma(x,\theta) = \left( x, \frac{\cos \theta}{x}, \frac{\sin \theta}{x} \right) \end{equation*}

The surface area of the Horn of Gabriel is quite interesting. (The comparison results for single-variable integrals are useful in this calculation.)

\begin{align*} \sigma_x \amp = \left( 1, \frac{-\cos \theta}{x^2}, \frac{-\sin \theta}{x^2} \right)\\ \sigma_{\theta} \amp = \left( 0, \frac{-\sin \theta}{x}, \frac{\cos \theta}{x} \right)\\ \sigma_x \times \sigma_{\theta} \amp = \left( \frac{-1}{x^3}, \frac{\cos \theta}{x}, \frac{-\sin \theta}{x} \right)\\ |\sigma_x \times \sigma_\theta| \amp = \sqrt{ \frac{1}{x^6} + \frac{1}{x^2}} = \frac{1}{x} \sqrt{ \frac{1}{x^4} + 1}\\ A \amp = \int_\sigma 1 = \int_1^{\infty} \int_0^{2\pi} \frac{1}{x} \sqrt{ \frac{1}{x^4} + 1} d\theta dx\\ \amp = 2\pi \int_1^{\infty} \frac{1}{x} \sqrt{ \frac{1}{x^4} + 1} dx \geq 2\pi \int_1^\infty \frac{1}{x} = \infty \end{align*}

Compare this surface are with the volume of the Horn of Gabriel.

\begin{align*} V \amp = \int_1^{\infty} \int_0^{\frac{1}{x}} \int_0^{2\pi} r d \theta dr dx\\ \amp = 2\pi \int_1^\infty \left. \frac{r^2}{2} \right|_0^{\frac{1}{x}} = 2\pi \int_1^\infty \frac{1}{2x^2} dx\\ \amp = \left. \frac{2\pi}{2} \frac{-1}{x} \right|_1^\infty = \pi \end{align*}

This is a very strange situation: an object with finite volume and infinite surface area.