Problem 26 - Thomas Calculus

Question

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Cone } \mathbf{F}=y^{2} \mathbf{i}+x z \mathbf{j}-\mathbf{k} \text { outward (normal away from the }} \\ {z \text { -axis through the cone } z=2 \sqrt{x^{2}+y^{2}}, 0 \leq z \leq 2}\end{array}\)

Step-by-Step Solution

Answer

The flux is \(-2\pi\).

Step 1: Parameterize the Surface

To parameterize the cone given by the equation \( z = 2 \sqrt{x^2 + y^2} \) in cylindrical coordinates, let \( r \) and \( \theta \) vary as follows: \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), \( z = 2r \). This gives the parametric representation \( \mathbf{r}(r, \theta) = (r \cos\theta, r \sin\theta, 2r) \) where \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \).

Step 2: Compute the Normal Vector

Find the normal vector \( \mathbf{n} \) by calculating the cross product of the partial derivatives \( \frac{\partial \mathbf{r}}{\partial r} = (\cos\theta, \sin\theta, 2) \) and \( \frac{\partial \mathbf{r}}{\partial \theta} = (-r \sin\theta, r \cos\theta, 0) \). Thus, \( \mathbf{n} = \frac{\partial \mathbf{r}}{\partial r} \times \frac{\partial \mathbf{r}}{\partial \theta} = (-2r\cos\theta, -2r\sin\theta, r) \). To ensure the outward direction, check and adjust the sign if necessary.

Step 3: Simplify the Dot Product \(\mathbf{F} \cdot \mathbf{n}\)

Substitute the parametrization into \( \mathbf{F} = y^2 \mathbf{i} + xz \mathbf{j} - \mathbf{k} \) to get \( \mathbf{F}(r, \theta) = (r^2 \sin^2\theta, 2r^2 \cos\theta \sin\theta, -1) \). The dot product \( \mathbf{F} \cdot \mathbf{n} = (r^2 \sin^2\theta)(-2r\cos\theta) + (2r^2 \cos\theta \sin\theta)(-2r\sin\theta) + (-1)(r) = -2r^3\sin^2\theta\cos\theta - 4r^3\cos\theta\sin^2\theta - r \).

Step 4: Set Up and Evaluate the Double Integral

The flux is given by the double integral \( \iint_{D} (\mathbf{F} \cdot \mathbf{n}) \, dA \) in polar coordinates. Substitute the expression from Step 3 to get \( \int_{0}^{2\pi} \int_{0}^{1} (-2r^3 \sin^2\theta \cos\theta - 4r^3 \cos\theta \sin^2\theta - r) \, r \, dr \, d\theta \). Simplify and calculate this integral to evaluate the flux.

Step 5: Evaluate Integrals Separately

Evaluate \( \int_{0}^{2\pi} \int_{0}^{1} (-2r^4 \sin^2\theta \cos\theta - 4r^4 \cos\theta \sin^2\theta) \, dr \, d\theta \) and \( \int_{0}^{2\pi} \int_{0}^{1} (-r^2) \, dr \, d\theta \). After integrating with respect to \( r \) and \( \theta \), find the sum of these integrals. Note that the symmetry of \( \sin^2\theta \cos\theta \) and \( \cos\theta \sin^2\theta \) across a full period will result in their integrals equating to zero.

Key Concepts

Parametrization

Parametrization is a method used to describe a surface using parameters, usually one or two variables, in terms of coordinates. In this exercise, we are working with a conical surface, which can be complex to describe using Cartesian coordinates. By using parametrization, particularly in cylindrical coordinates, we simplify the surface description.

In parametrization, we express the coordinates of the surface as functions of these parameters. For a cone, parametrized representations are typically chosen because it aligns with the symmetry of the problem. This means we tie together radius and angle in the plane with height along the axis. It's providing an efficient description through variables such as \(r\) for radius and \(\theta\) for angle.

• This approach converts the 3D problem into a form that is easier to integrate due to the regularity in shapes like circles and surfaces like cones.
• In our specific problem, the cone is defined above the \(xy\)-plane by \(z = 2\sqrt{x^2 + y^2}\).

By assigning the parameters \(x = r\cos(\theta), y = r\sin(\theta), z = 2r\), we efficiently describe all points on the cone with a simple range for \(r \) and \( \theta \). This makes further calculation, such as finding the flux, much more systematic.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates into three dimensions with an additional height parameter. In this case, the coordinates are represented as \( (r, \theta, z) \), where:

• \( r \) is the distance from the z-axis, similar to the radius in polar coordinates.
• \( \theta \) is the angle counterclockwise from the positive x-axis, also akin to polar coordinates.
• \( z \) is the height above the \( xy \)-plane.

This system is handy for problems with symmetry around an axis, such as cones or cylinders. By transforming the problem from Cartesian to cylindrical, we naturally align with the geometry of these objects.

In our cone example, transforming to cylindrical coordinates allowed us to change a complex conical surface into a much simpler form. The given cone is identified by its transformation \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( z = 2r \) which aligns with the nature of cones having circular base symmetry.

This makes setting up and solving integrals easier because the limits and calculations are more straightforward. We operate over a consistent range of \( r \) (0 to 1) and \( \theta \) (0 to \( 2\pi \)), instead of varying limits that would appear in Cartesian coordinates.

Double Integral

A double integral is a tool used to calculate values over a region in a two-dimensional plane. In the context of surface flux, double integrals are used to integrate over surfaces. It allows us to sum up infinitesimal contributions from the surface area to find, for example, total flux passing through a surface.

When using cylindrical coordinates for a double integral, it's crucial to recall that the differential area element \(dA\) becomes \( r\, dr\, d\theta \).

For our surface integral over a cone, you substitute the surface parametrization into the vector field function and compute the normal vector's dot product with the field. This results in an expression which is then integrated over the specified range. The integration essentially adds up the vector field influences multiplied by the area element to find a total effect across the cone’s surface area.

• This involves simplifying the expression before integrating — making sure that terms are correctly aligned to receive simple integration limits.

The cone problem showcases how cylindrical symmetry simplifies integration; the \(r\, dr\, d\theta \) accounts for the circular sections inherent to the cone’s shape and facilitates solving the double integral efficiently.

Normal Vector

A normal vector to a surface is a vector that is perpendicular to every tangent plane of that surface at a given point, and it plays a crucial role in aspects like surface integration and determining directionality of phenomena such as flux.

In the context of flux computation, finding the correct orientation of the normal vector is essential as it dictates whether the calculated flux is inward or outward. Normal vectors are obtained through the cross product of two tangent vectors to the surface.

• These tangent vectors come from the derivatives of the parametrization with respect to each parameter.
• For the parameterized cone, we derive \( \frac{\partial \mathbf{r}}{\partial r} \) and \( \frac{\partial \mathbf{r}}{\partial \theta} \) as the tangent vectors.

The normal vector \( \mathbf{n} \) is obtained from their cross product and for our parameterization is \((-2r \cos \theta, -2r \sin \theta, r)\). Correcting the direction is done by checking the vector's components' alignment, ensuring it points outward as required in our problem definition.
In solving such exercises, verifying the sign and direction is vital, potentially involving additional orientation checks to ascertain outward flux (i.e., away from the \(z\)-axis for the conical surface).