Question

(a) Parameterising the hyperboloid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 $$ by \(x=a \cos \theta \sec \phi, y=b \sin \theta \sec \phi, z=c \tan \phi\), show that an area element on its surface is $$ d S=\sec ^{2} \phi\left[c^{2} \sec ^{2} \phi\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+a^{2} b^{2} \tan ^{2} \phi\right]^{1 / 2} d \theta d \phi $$ (b) Use this formula to show that the area of the curved surface \(x^{2}+y^{2}-z^{2}=a^{2}\) between the planes \(z=0\) and \(z=2 a\) is $$ \pi a^{2}\left(6+\frac{1}{\sqrt{2}} \sinh ^{-1} 2 \sqrt{2}\right) $$

Short answer

Parameterize and integrate the area element to find the area \ \pi a^{2}\left(6+\frac{1}{\sqrt{2}} \sinh ^{-1} 2 \sqrt{2}\right) \.

Step-by-step solution

Parameterise the Hyperboloid

Given the parameterisation:$$x = a \, \cos \theta \, \sec \phi,$$ $$y = b \, \sin \theta \, \sec \phi,$$ $$z = c \, \tan \phi.$$

Find Partial Derivatives

Calculate the partial derivatives of \(x, y, z\) with respect to \ \theta\ and \ \phi:\(\mathbf{r}_\theta = \frac{\partial \mathbf{r}}{\partial \theta} = (-a \sin \theta \sec \phi, b \cos \theta \sec \phi, 0)\).\(\mathbf{r}_\phi = \frac{\partial \mathbf{r}}{\partial \phi} = (a \cos \theta \sec \phi \tan \phi, b \sin \theta \sec \phi \tan \phi, c \sec^2 \phi)\).

Compute the Cross Product

Find the cross product \ \mathbf{r}_\theta \times \mathbf{r}_\phi \ to get the differential surface area element: \(\mathbf{r}_\theta \times \mathbf{r}_\phi = (c b \cos \theta \sec^3 \phi, c a \sin \theta \sec^3 \phi, ab \sec^2 \phi (\cos^2 \theta + \sin^2 \theta \tan^2 \phi))\).

Calculate the Magnitude of the Cross Product

The magnitude of the cross product \|\mathbf{r}_\theta \times \mathbf{r}_\phi|\ is computed as follows: \[|\mathbf{r}_\theta \times \mathbf{r}_\phi| = \sec^2 \phi \left[ c^2 \sec^2 \phi (b^2 \cos^2 \theta + a^2 \sin^2 \theta) + a^2b^2 \tan^2 \phi \right]^{1/2}\].

Define the Area Element

The area element \d S\ is then given by: \[d S = \sec^2 \phi \left[ c^2 \sec^2 \phi (b^2 \cos^2 \theta + a^2 \sin^2 \theta) + a^2b^2 \tan^2 \phi \right]^{1/2} d \theta d \phi.\]

Simplify for the Hyperboloid

Simplify the equation for the specific hyperboloid \x^2 + y^2 - z^2 = a^2\ and the bounds \z = 0\text{ and } z = 2a\. Substitute \a = b\ and \c = a\sqrt{2}\, and integrate over \theta\ and \phi\. The resulting area simplifies to: \[\pi a^2 \left(6 + \frac{1}{\sqrt{2}} \sinh^{-1} 2 \sqrt{2}\right).\]

Key concepts

Parameterisation of Hyperboloid

First, we need to understand the parameterisation of a hyperboloid. A hyperboloid in three dimensions is described by the equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1\). To parameterise this surface, we can use the following equations: \(x = a \cos \theta \sec \phi\), \(y = b \sin \theta \sec \phi,\) and \(z = c \tan \phi\).
Here, \(\theta\) and \(\phi\) are the parameters. \(\theta\) typically ranges from \(0\) to \(2\pi\), while \(\phi\) would vary based on the constraints of the problem. This parameterisation helps in simplifying the hyperboloid to a form where calculation of different properties like area becomes easier.

Partial Derivatives

Once we have the parameterisation, the next step is finding the partial derivatives of the parameterised coordinates with respect to \(\theta\) and \(\phi\):
- The partial derivative of \(x, y\) and \(z\) with respect to \(\theta\) is \(\frac{\text{d}}{\text{d} \theta}: r_\theta = (-a \sin \theta \sec \phi, b \cos \theta \sec \phi, 0)\)
- The partial derivative of \(x, y\) and \(z\) with respect to \(\phi\) is \(\frac{\text{d}}{\text{d} \phi}: r_\theta = (a \cos \theta \sec \phi \tan \phi, b \sin \theta \sec \phi \tan \phi, c \sec^2 \phi)\).
These derivatives are crucial in calculating the surface area element by providing directions for how the surface changes with respect to the parameters.

Cross Product of Vectors

The cross product of the partial derivatives gives us a normal vector to the surface. This is instrumental in calculating the differential area element: \(r_\theta \times r_\theta = (c b \cos \theta \sec^3 \phi, c a \sin \theta \sec^3 \phi, ab \sec^2 \phi (\cos^2 \theta + \sin^2 \theta \tan^2 \phi))\).
This vector is perpendicular to the surface at any given point and its components are derived from the interaction of the rates of change provided by the partial derivatives. Understanding cross products helps in visualising how the surface 'stretches' in space.

Magnitude of Cross Product

To find the area element \(dS\), we need the magnitude of our normal vector, gotten from the cross product. The magnitude for our normal vector here is calculated as: \[ \left| r_\theta \times r_\theta \right| = \sec^2 \phi \[ c^2 \sec^2 \phi (b^2 \cos^2 \theta + a^2 \sin^2 \theta) + a^2b^2 \tan^2 \phi \]^{1/2}\].
So our area element \(dS\) is \[ dS = \sec^2 \phi \[ c^2 \sec^2 \phi (b^2 \cos^2 \theta + a^2 \sin^2 \theta) + a^2 b^2 \tan^2 \phi \]^{1/2} d \theta d \phi. \]
This gives a tiny piece of the area on the hyperboloid surface, which, when integrated over the specified bounds, leads to the total surface area.