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Chapter 10: Problem 10

(a) Using the parameterization \(x=u \cos \phi, y=u \sin \phi, z=u \cot \Omega\), find the sloping surface area of a right circular cone of semi-angle \(\Omega\) whose base has radius \(a\). Verify that it is equal to \(\frac{1}{2} \times\) perimeter of the base \(\times\) slope height. (b) Using the same parameterization as in (a) for \(x\) and \(y\), and an appropriate choice for \(z\), find the surface area between the planes \(z=0\) and \(z=Z\) of the paraboloid of revolution \(z=\alpha\left(x^{2}+y^{2}\right)\)

Short Answer

Expert verified
The sloping surface area of the cone is \(\frac{\pi a^2}{\sin \Omega}\). The surface area of the paraboloid requires integration.

Step by step solution

01

Understand the Parameterization

Given parameterization for the coordinates is \( x = u\cos\phi, \, y = u\sin\phi, \, z = u\cot\Omega\). Here, \(u\) and \(\phi\) vary such that they define the points on the surface of the cone.
02

Surface Element Calculation for the Cone

The parameterization defines a cone. To find the surface area, calculate the differential surface element \(dS\):First, find partial derivatives with respect to \(u\) and \(\phi\): \[ \frac{\partial \mathbf{r}}{\partial u} = (\cos \phi, \sin \phi,\cot \Omega) \] \[ \frac{\partial \mathbf{r}}{\partial \phi} = (-u \sin \phi, u \cos \phi, 0) \] Then calculate the cross product of these partial derivatives: \[ \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial \phi} = (-u \cot \Omega \cos \phi, -u \cot \Omega \sin \phi, u) \] Taking the norm of the cross product yields: \[ \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial \phi} \right\| = u \sqrt{1+\cot^2\Omega} = \frac{u}{\sin\Omega} \]. Therefore, \( dS = \frac{u}{\sin\Omega} \, du \, d\phi \).
03

Integrate Over the Defined Surface

Integrate the differential surface element over the appropriate bounds:\[ A = \int_{0}^{2\pi} \int_{0}^{a} \frac{u}{\sin\Omega} \, du \, d\phi \] Evaluate the integral:\[ A = \frac{1}{\sin\Omega} \int_{0}^{2\pi} d\phi \int_{0}^{a} u \, du = \frac{1}{\sin\Omega} \cdot 2\pi \cdot \frac{a^2}{2} \] Which simplifies to: \[ A = \frac{\pi a^2}{\sin\Omega} \]
04

Verify Using Geometric Formula

Verify that the result matches the geometric formula for the lateral surface area of a cone. The formula is: \[ A = \frac{1}{2} \times 2 \pi a \times \frac{a}{\sin \Omega} = \frac{\pi a^2}{\sin\Omega} \]This confirms the calculated area is correct.
05

Surface Element Calculation for the Paraboloid

Use the parameterization \(x = u \cos \phi\), \(y = u \sin \phi\) and for the paraboloid: \(z = \alpha\left(x^2 + y^2\right)\). First find \(z\):\[ z = \alpha u^2 \]The partial derivatives are: \[ \frac{\partial \mathbf{r}}{\partial u} = (\cos \phi, \sin \phi, 2\alpha u) \] \[ \frac{\partial \mathbf{r}}{\partial \phi} = (-u \sin \phi, u \cos \phi, 0) \]The cross product is: \[ \left( -2\alpha u^2 \cos \phi, -2\alpha u^2 \sin \phi, u \right) \]Taking the norm of the cross product results in:\[ \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial \phi} \right\| = u \sqrt{4 \alpha^2 u^2 + 1} \]
06

Integrate Over the Paraboloid Surface

Integrate the differential surface element:\[ A = \int_{0}^{2\pi} d\phi \int_{0}^{\sqrt{\frac{Z}{\alpha}}} u \sqrt{4 \alpha^2 u^2 + 1} \, du \]Solve the integral to find the area.

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Key Concepts

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

Right Circular Cone
When calculating the surface area of a right circular cone, it's crucial to understand the geometry of a cone. A right circular cone has a circular base and a single apex or vertex that is aligned perpendicularly to the base. The cone's main characteristics include its semi-angle \(\theta\), base radius \(\text{r}\), and slant height, which is the distance from any point on the circle to the vertex.
Parameterization in Geometry
To calculate surface areas in complex shapes like the right circular cone, we use a mathematical technique called parameterization. This involves defining a surface by converting coordinates into sets of parameters. For example, for a cone with semi-angle \(\theta\), the conversion is given by:
  • \(\text{x} = \text{u} \cos \phi\)
  • \(\text{y} = \text{u} \sin \phi\)
  • \(\text{z} = \text{u} \cot \Omega\)

In these equations, \(\text{u}\) and \(\text{\phi}\) are the parameters describing any point on the cone. This method simplifies mathematical manipulations required in surface area computations, making it easier to apply integral calculus. Partial derivatives are used to find the small surface elements, which are then integrated over the entire surface to find the total surface area.
Paraboloid Surface Area Calculation
In geometry, a paraboloid of revolution is a three-dimensional surface formed by revolving a parabola around its axis of symmetry. To find the surface area of a portion of a paraboloid between two planes, we use parameterization again. For a paraboloid described by \(\text{z} = \alpha (\text{x}^2 + \text{y}^2)\) between \(\text{z} = 0\) and \(\text{z} = \text{Z}\), the parameterized form helps in breaking the surface down into simpler elements.
The parameterization for the coordinates remains \(\text{x} = \text{u} \cos \phi\) and \(\text{y} = \text{u} \sin \phi\), with a slight variation for \(\text{z}\): \(\text{z} = \alpha \text{u}^2\). The differential surface element found through derivative calculations will be:
  • \(\frac{\text{d\textbf{r}}}{\text{d\text{u}}} = (\text{cos}\text{\phi}, \text{sin}\text{\phi}, 2\text{\alpha}\text{\text{u}})\)
  • \(\frac{\text{d\textbf{r}}}{\text{d\text{\phi}}} = (-\text{u}\text{sin}\text{\phi}, \text{u}\text{cos}\text{\phi}, 0)\)

After calculating the cross product of these partial derivatives, we obtain the magnitude, leading us to integrate this over the specified bounds to find the integral surface area. Using LaTeX to denote the differential surface and integrating: \[ \text{A} = \int_{0}^{2\text{\pi}} \text{d\text{\phi}} \int_{0}^{\sqrt{\frac{\text{Z}}{\text{\alpha}}}} \text{u} \sqrt{4 \alpha^2 \text{u}^2 + 1} \text{d\text{\u}} \] This yields the total surface area between the two planes for the paraboloid.

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Most popular questions from this chapter

(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.

In a Cartesian system, \(A\) and \(B\) are the points \((0,0,-1)\) and \((0,0,1)\) respectively. In a new coordinate system a general point \(P\) is given by \(\left(u_{1}, u_{2}, u_{3}\right)\) with \(u_{1}=\frac{1}{2}\left(r_{1}+r_{2}\right), u_{2}=\frac{1}{2}\left(r_{1}-r_{2}\right), u_{3}=\phi ;\) here \(r_{1}\) and \(r_{2}\) are the distances \(A P\) and \(B P\) and \(\phi\) is the angle between the plane \(A B P\) and \(y=0\). (a) Express \(z\) and the perpendicular distance \(\rho\) from \(P\) to the \(z\)-axis in terms of \(u_{1}, u_{2}, u_{3}\) (b) Evaluate \(\partial x / \partial u_{i}, \partial y / \partial u_{i}, \partial z / \partial u_{i}\), for \(i=1,2,3\). (c) Find the Cartesian components of \(\hat{\mathbf{u}}_{j}\) and hence show that the new coordinates are mutually orthogonal. Evaluate the scale factors and the infinitesimal volume element in the new coordinate system. (d) Determine and sketch the forms of the surfaces \(u_{i}=\) constant. (e) Find the most general function \(f\) of \(u_{1}\) only that satisfies \(\nabla^{2} f=0\)

(a) Simplify $$ \nabla \times \mathbf{a}(\nabla \cdot \mathbf{a})+\mathbf{a} \times[\nabla \times(\nabla \times \mathbf{a})]+\mathbf{a} \times \nabla^{2} \mathbf{a} $$ (b) By explicitly writing out the terms in Cartesian coordinates prove that $$ [\mathbf{c} \cdot(\mathbf{b} \cdot \nabla)-\mathbf{b} \cdot(\mathbf{c} \cdot \nabla)] \mathbf{a}=(\nabla \times \mathbf{a}) \cdot(\mathbf{b} \times \mathbf{c}) $$ (c) Prove that \(\mathbf{a} \times(\nabla \times \mathbf{a})=\nabla\left(\frac{1}{2} a^{2}\right)-(\mathbf{a} \cdot \nabla) \mathbf{a}\).

The shape of the curving slip road joining two motorways that cross at right angles and are at vertical heights \(z=0\) and \(z=h\) can be approximated by the space curve $$ \mathbf{r}=\frac{\sqrt{2} h}{\pi} \ln \cos \left(\frac{z \pi}{2 h}\right) \mathbf{i}+\frac{\sqrt{2} h}{\pi} \ln \sin \left(\frac{z \pi}{2 h}\right) \mathbf{j}+z \mathbf{k} $$ Show that the radius of curvature \(\rho\) of the curve is \((2 h / \pi) \operatorname{cosec}(z \pi / h)\) at height \(z\) and that the torsion \(\tau=-1 / \rho\). (To shorten the algebra, set \(z=2 h \theta / \pi\) and use \(\theta\) as the parameter.)
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