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Chapter 10: Problem 11

(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

Expert verified
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

01

Parameterise the Hyperboloid

Given the parameterisation:$$x = a \, \cos \theta \, \sec \phi,$$ $$y = b \, \sin \theta \, \sec \phi,$$ $$z = c \, \tan \phi.$$
02

Find Partial Derivatives

Calculate the partial derivatives of \(x, y, z\) with respect to \ \theta\ and \ \phi:\(\mathbf{r}_\theta = \frac{\partial \mathbf{r}}{\partial \theta} = (-a \sin \theta \sec \phi, b \cos \theta \sec \phi, 0)\).\(\mathbf{r}_\phi = \frac{\partial \mathbf{r}}{\partial \phi} = (a \cos \theta \sec \phi \tan \phi, b \sin \theta \sec \phi \tan \phi, c \sec^2 \phi)\).
03

Compute the Cross Product

Find the cross product \ \mathbf{r}_\theta \times \mathbf{r}_\phi \ to get the differential surface area element: \(\mathbf{r}_\theta \times \mathbf{r}_\phi = (c b \cos \theta \sec^3 \phi, c a \sin \theta \sec^3 \phi, ab \sec^2 \phi (\cos^2 \theta + \sin^2 \theta \tan^2 \phi))\).
04

Calculate the Magnitude of the Cross Product

The magnitude of the cross product \|\mathbf{r}_\theta \times \mathbf{r}_\phi|\ is computed as follows: \[|\mathbf{r}_\theta \times \mathbf{r}_\phi| = \sec^2 \phi \left[ c^2 \sec^2 \phi (b^2 \cos^2 \theta + a^2 \sin^2 \theta) + a^2b^2 \tan^2 \phi \right]^{1/2}\].
05

Define the Area Element

The area element \d S\ is then given by: \[d S = \sec^2 \phi \left[ c^2 \sec^2 \phi (b^2 \cos^2 \theta + a^2 \sin^2 \theta) + a^2b^2 \tan^2 \phi \right]^{1/2} d \theta d \phi.\]
06

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: \[\pi a^2 \left(6 + \frac{1}{\sqrt{2}} \sinh^{-1} 2 \sqrt{2}\right).\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Maxwell's equations for electromagnetism in free space (i.e. in the absence of charges, currents and dielectric or magnetic media) can be written (i) \(\nabla \cdot \mathbf{B}=0\) (ii) \(\nabla \cdot \mathbf{E}=0\) (iii) \(\nabla \times \mathbf{E}+\frac{\partial \mathbf{B}}{\partial t}=\mathbf{0}\) (iv) \(\nabla \times \mathbf{B}-\frac{1}{c^{2}} \frac{\partial \mathbf{E}}{\partial t}=\mathbf{0}\). A vector \(\mathbf{A}\) is defined by \(\mathbf{B}=\nabla \times \mathbf{A}\), and a scalar \(\phi\) by \(\mathbf{E}=-\nabla \phi-\partial \mathbf{A} / \partial t\). Show that if the condition (v) \(\nabla \cdot \mathbf{A}+\frac{1}{c^{2}} \frac{\partial \phi}{\partial t}=0\) is imposed (this is known as choosing the Lorenz gauge), then both \(\mathbf{A}\) and \(\phi\) satisfy the wave equations (vi) \(\nabla^{2} \phi-\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}=0\), (vii) \(\quad \nabla^{2} \mathbf{A}-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}}=\mathbf{0}\) The reader is invited to proceed as follows. (a) Verify that the expressions for \(\mathbf{B}\) and \(\mathbf{E}\) in terms of \(\mathbf{A}\) and \(\phi\) are consistent with (i) and (iii). (b) Substitute for \(\mathbf{E}\) in (ii) and use the derivative with respect to time of \((v)\) to eliminate \(\mathbf{A}\) from the resulting expression. Hence obtain (vi). (c) Substitute for \(\mathbf{B}\) and \(\mathbf{E}\) in (iv) in terms of \(\mathbf{A}\) and \(\phi .\) Then use the divergence of (v) to simplify the resulting equation and so obtain (vii).

In a magnetic field, field lines are curves to which the magnetic induction \(\mathbf{B}\) is everywhere tangential. By evaluating \(d \mathbf{B} / d s\), where \(s\) is the distance measured along a field line, prove that the radius of curvature at any point on a line is given by $$ \rho=\frac{B^{3}}{|\mathbf{B} \times(\mathbf{B} \cdot \nabla) \mathbf{B}|} $$

(a) For cylindrical polar coordinates \(\rho, \phi, z\) evaluate the derivatives of the three unit vectors with respect to each of the coordinates, showing that only \(\partial \hat{\mathbf{e}}_{\rho} / \partial \phi\) and \(\partial \hat{\mathbf{e}}_{\phi} / \partial \phi\) are non-zero. (i) Hence evaluate \(\nabla^{2} \mathbf{a}\) when \(\mathbf{a}\) is the vector \(\hat{\mathbf{e}}_{\rho}\), i.e. a vector of unit magnitude everywhere directed radially outwards from the \(z\)-axis. (ii) Note that it is trivially obvious that \(\nabla \times \mathbf{a}=\mathbf{0}\) and hence that equation \((10.41)\) requires that \(\dot{\nabla}(\nabla \cdot \mathbf{a})=\nabla^{2} \mathbf{a}\). (iii) Evaluate \(\nabla(\nabla \cdot \mathbf{a})\) and show that the latter equation holds, but that $$ [\nabla(\nabla \cdot \mathbf{a})]_{\rho} \neq \nabla^{2} a_{\rho} $$ (b) Rework the same problem in Cartesian coordinates (where, as it happens, the algebra is more complicated).

At time \(t=0\), the vectors \(\mathbf{E}\) and \(\mathbf{B}\) are given by \(\mathbf{E}=\mathbf{E}_{0}\) and \(\mathbf{B}=\mathbf{B}_{0}\), where the fixed unit vectors \(\mathbf{E}_{0}\) and \(\mathbf{B}_{0}\) are orthogonal. The equations of motion are $$ \begin{aligned} &\frac{d \mathbf{E}}{d t}=\mathbf{E}_{0}+\mathbf{B} \times \mathbf{E}_{0} \\ &\frac{d \mathbf{B}}{d t}=\mathbf{B}_{0}+\mathbf{E} \times \mathbf{B}_{0} \end{aligned} $$ Find \(\mathbf{E}\) and \(\mathbf{B}\) at a general time \(t\), showing that after a long time the directions of \(\mathbf{E}\) and \(\mathbf{B}\) have almost interchanged.

Evaluate the Laplacian of the function $$ \psi(x, y, z)=\frac{z x^{2}}{x^{2}+y^{2}+z^{2}} $$ (a) directly in Cartesian coordinates, and (b) after changing to a spherical polar coordinate system. Verify that, as they must, the two methods give the same result.
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