Question

(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

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

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.

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, \( dS = \frac{u}{\sin\Omega} \, du \, d\phi \).

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: \[ A = \frac{\pi a^2}{\sin\Omega} \]

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.

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:\[ \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial \phi} \right\| = u \sqrt{4 \alpha^2 u^2 + 1} \]

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.

Key concepts

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.