Problem 381 - Calculus Volume 3

Question

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral \(\int_{S} \mathbf{F} \cdot \mathbf{n} d s\) for the given choice of \(\mathbf{F}\) and the boundary surface \(S\). For each closed surface, assume \(\mathbf{N}\) is the outward unit normal vector. \(\quad\) [T] \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+\left(z^{2}-1\right) \mathbf{k} ; \quad S\) is the surface of the solid bounded by cylinder \(x^{2}+y^{2}=4\) and planes \(z=0\) and \(z=1\)

Step-by-Step Solution

Answer

The surface integral is \(12\pi\).

Step 1: Understand the Problem

The task is to evaluate the surface integral \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds \) over the surface \( S \), where \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + (z^2 - 1) \mathbf{k} \). The surface \( S \) is the boundary of the cylinder defined by \( x^2 + y^2 = 4 \) with planes \( z = 0 \) and \( z = 1 \). This forms a cylindrical region with top and bottom circular bases.

Step 2: Apply the Divergence Theorem

The Divergence Theorem relates the surface integral \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds \) to a volume integral over the region \( V \) enclosed by \( S \). It states that \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds = \int_{V} abla \cdot \mathbf{F} \, dV \). Thus, we need to compute the divergence of \( \mathbf{F} \), \( abla \cdot \mathbf{F} \).

Step 3: Compute the Divergence of \( \mathbf{F} \)

The divergence of \( \mathbf{F} \) is calculated as \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z^2 - 1) \). Simplifying, we get \( abla \cdot \mathbf{F} = 1 + 1 + 2z = 2 + 2z \).

Step 4: Setup the Volume Integral

The integral becomes \( \int_{V} (2 + 2z) \, dV \), where \( V \) is the volume of the cylinder bounded by \( x^2 + y^2 = 4 \), \( z = 0 \), and \( z = 1 \). This cylindrical region can be evaluated using cylindrical coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \), \( z = z \), with bounds \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq 1 \).

Step 5: Evaluate the Volume Integral

Convert the volume integral to cylindrical coordinates: \( \int_{0}^{2} \int_{0}^{2\pi} \int_{0}^{1} (2 + 2z) \, r \, dz \, d\theta \, dr \). The integral becomes: 1. \( \int_{0}^{1} (2 + 2z) \, dz = [2z + z^2]_{0}^{1} = 3 \).2. \( \int_{0}^{2\pi} d\theta = 2\pi \).3. \( \int_{0}^{2} r \, dr = [\frac{r^2}{2}]_{0}^{2} = 2 \).Calculate the product of these results: \( 3 \times 2\pi \times 2 = 12\pi \).

Step 6: Conclude the Calculation

The value of the surface integral \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, ds \) using the divergence theorem is given by the volume integral result, which equals \( 12\pi \).

Key Concepts

Surface Integral

A surface integral is a way to calculate a quantity over a two-dimensional surface. Imagine it as adding up a function's value over a whole surface, rather like adding up the height of a curve over a length. In our problem, we're dealing with a vector field \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + (z^2 - 1) \mathbf{k} \) and the surface \( S \) of a cylinder capped by two circular discs. The surface integral in this context is used to calculate the "flux" of a vector field through a surface, the total "flow" of the vector field through \( S \).

The Divergence Theorem provides a powerful tool to transform this surface integral into a possibly simpler volume integral. It states that the flux through a closed surface can be calculated by integrating the divergence of the field over the volume it encloses. This is particularly useful in complex geometries where direct computation of surface integrals might be hard. Utilizing the Divergence Theorem, as shown in the solution, involves computing the divergence \( abla \cdot \mathbf{F} \) and transforming the problem into an integration over a volume, which might present a simplification in computing the desired quantity.

Cylindrical Coordinates

Cylindrical coordinates offer a valuable way to evaluate integrals in geometries with circular symmetry, like cylinders. They relate to the Cartesian coordinates \( (x, y, z) \) through:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( z = z \)

Here, \( r \) is the radial distance from the center (like the hypotenuse in polar coordinates), \( \theta \) is the angle around the z-axis (much like in polar coordinates in two dimensions), and \( z \) represents the height.

In our exercise, where the cylinder's equation is \( x^2 + y^2 = 4 \), employing cylindrical coordinates makes it easier to state the problem's bounds for integration. This simplifies complex triple integrals by aligning them with the natural symmetries of a cylinder.

In terms of computing the volume integral in the solution, cylindrical coordinates enabled setting clear limits: \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq 1 \). The clarity and symmetry provided by these coordinates are ideal for problems involving circles or cylinders, aiding in moving efficiently through calculations.

Volume Integral

A volume integral sums up a function throughout a volume, adding up values inside a 3D space. In the context of the Divergence Theorem, we leverage volume integrals to convert what would be a complex surface calculation into a more manageable form. By using the divergence \( abla \cdot \mathbf{F} \), the volume integral component of \( \int_{V} (2 + 2z) \, dV \) within the enclosed volume of the cylinder can be computed.

In our solved example, using cylindrical coordinates allowed us to structure the integral clearly:\begin{align*}\int_{0}^{2} \int_{0}^{2\pi} \int_{0}^{1} (2 + 2z) \, r \, dz \, d\theta \, dr.\end{align*}This resulted in sequential integrations with respect to \( z \), \( \theta \), and \( r \). For each of these variables, the contributions were calculated:

• \( \int_{0}^{1} (2 + 2z) \, dz = 3 \) representing the contribution of height in the cylinder.
• \( \int_{0}^{2\pi} d\theta = 2\pi \) representing the full rotation around the cylinder.
• \( \int_{0}^{2} r \, dr = 2 \) representing the radial direction from the center to the boundary.

The result of \( 3 \times 2\pi \times 2 = 12\pi \) provided the final answer, showing that breaking down integrals using suitable coordinates and methods simplifies calculations significantly.