Problem 19 - Thomas Calculus

Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Cone frustum The portion of the cone \(z=2 \sqrt{x^{2}+y^{2}}\) between the planes \(z=2\) and \(z=6\)

Step-by-Step Solution

Answer

The area of the cone frustum is \(8\pi\sqrt{5}\).

Step 1: Understand the Cone's Surface

The equation given is for a cone, where \( z = 2 \sqrt{x^2 + y^2} \). This describes a cone with its apex at the origin and opening upward. The cone is truncated between the planes \( z = 2 \) and \( z = 6 \). We need to find the area of this frustum.

Step 2: Parametrize the Cone Surface

To parametrize the surface, convert the cone equation to cylindrical coordinates: \( x = r\cos\theta \), \( y = r\sin\theta \), and \( z = 2r \). The variable \( r \) represents the radius, which ranges from \( 1 \) to \( 3 \) because for \( z = 2 \), \( r = 1 \), and for \( z = 6 \), \( r = 3 \). \( \theta \) ranges from \( 0 \) to \( 2\pi \). Therefore, the parametrization becomes \( \mathbf{r}(r, \theta) = (r\cos\theta, r\sin\theta, 2r) \).

Step 3: Calculate the Surface Elements

To find the surface elements, calculate the partial derivatives: \[ \mathbf{r}_r = (\cos\theta, \sin\theta, 2) \] \[ \mathbf{r}_\theta = (-r\sin\theta, r\cos\theta, 0) \]Then, calculate the cross product \( \mathbf{r}_r \times \mathbf{r}_\theta \) to obtain the normal vector: \[ \mathbf{r}_r \times \mathbf{r}_\theta = (-2r\cos\theta, -2r\sin\theta, r) \] The magnitude of this vector gives the differential area element \( dS \):\[ ||\mathbf{r}_r \times \mathbf{r}_\theta|| = \sqrt{(-2r\cos\theta)^2 + (-2r\sin\theta)^2 + r^2} = \sqrt{5r^2} = \sqrt{5}r \] Therefore, \( dS = \sqrt{5}r \, dr \, d\theta \).

Step 4: Set Up the Double Integral

The area \( A \) of the frustum is given by the double integral over the surface:\[ A = \int_0^{2\pi} \int_1^3 \sqrt{5}r \, dr \, d\theta \]

Step 5: Evaluate the Double Integral

First, evaluate the inner integral with respect to \( r \):\[ \int_1^3 r \, dr = \left[\frac{r^2}{2} \right]_1^3 = \frac{9}{2} - \frac{1}{2} = 4 \]Then, evaluate the outer integral with respect to \( \theta \):\[ A = \int_0^{2\pi} \sqrt{5} \times 4 \, d\theta = 4\sqrt{5} \left[ \theta \right]_0^{2\pi} = 8\pi\sqrt{5} \]

Step 6: Finalize Your Area Calculation

The calculated area of the frustum is \( 8\pi\sqrt{5} \).

Key Concepts

Parametrization

Parametrization is a powerful tool in calculus, which allows us to express complex surfaces and curves in a manageable form. For a 3D shape like a cone or a frustum (a cone with the top cut off), parametrization lets us define every point on the surface using parameters. Instead of traditional Cartesian coordinates (x, y, z), we use parameters that can simplify integration and other calculations.

For example, given the equation of a cone, like \( z = 2 \sqrt{x^2 + y^2} \), the parametrization process helps convert this into a form suitable for integration. In surface area calculations, this often involves expressing the coordinates in terms of "radius" and "angle," as we did by introducing \( r \) and \( \theta \) in cylindrical coordinates. This conversion captures the essence of the surface in a way that aligns with its geometric properties.

By setting up the vector \( \mathbf{r}(r, \theta) = (r\cos\theta, r\sin\theta, 2r) \), the surface is navigated using \( r \) ranging from 1 to 3, and \( \theta \) from 0 to \( 2\pi \). This method ensures that each point on the truncated cone surface is uniquely identified by \( r \) and \( \theta \).

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that is particularly useful in dealing with problems exhibiting cylindrical symmetry. They extend polar coordinates by adding height as a third dimension, making it ideal for situations like analyzing a cone or circular cylinders.

In cylindrical coordinates, any point in space is represented by three parameters: \( (r, \theta, z) \), where:

• \( r \): the radial distance from the z-axis (also understood as the radius of the circular component)
• \( \theta \): the angular coordinate (angle measured from the positive x-axis)
• \( z \): the height in the vertical z-direction

For the given exercise, the cone's shape suggests using cylindrical coordinates because its equation naturally aligns with these parameters. Transforming \( x = r\cos\theta \) and \( y = r\sin\theta \), and since the equation takes the form \( z = 2r \), it becomes quite straightforward to integrate over these parameters.

This representation not only simplifies the calculation process but also helps visualize how each slice of the frustum integrates into the total surface area. It’s like taking horizontal slices and stacking them up to form the entire structure.

Vector Calculus

Vector calculus is a branch of mathematics focused on calculating the properties of vector fields, and it's crucial for dealing with multidimensional spaces, like those encountered in surface area problems. It includes operations like differentiation, integration, and finding vector products, which are essential in determining quantities such as flux and surface area.

To find the surface area of a truncated cone using vector calculus, we first derive the partial derivatives of the parametrization. These derivatives capture the "direction" of the surface elements at each point. For instance, \( \mathbf{r}_r = (\cos\theta, \sin\theta, 2) \) and \( \mathbf{r}_\theta = (-r\sin\theta, r\cos\theta, 0) \) describe how the surface changes with respect to \( r \) and \( \theta \).

Next, to find the differential area element, we compute the cross product \( \mathbf{r}_r \times \mathbf{r}_\theta \). This operation results in a vector that is perpendicular to the surface at each point, and its magnitude \( ||\mathbf{r}_r \times \mathbf{r}_\theta|| = \sqrt{5}r \) gives the area of a small patch on the surface.

Finally, integrating this area element over the specified ranges \( \int_1^3 \int_0^{2\pi} \sqrt{5}r \, dr \, d\theta \) yields the total surface area we are interested in. This method leverages the power of vector fields to simplify and accurately compute complex geometrical properties.