Problem 383 - 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. \(\mathbf{F}(x, y, z)=x^{3} \mathbf{i}+y^{3} \mathbf{j}+3 a^{2} z \mathbf{k}(\) constant \(a>0) ; \quad S\) is the surface bounded by cylinder \(x^{2}+y^{2}=a^{2}\) and planes \(z=0\) and \(z=1\)

Step-by-Step Solution

Answer

The surface integral is \(3\pi a^4\).

Step 1: Understand the Surface and Vector Field

The surface \(S\) is the boundary of a cylinder oriented along the \(z\)-axis with base radius \(a\), constrained between \(z=0\) and \(z=1\). The vector field is \(\mathbf{F}(x, y, z) = x^3 \mathbf{i} + y^3 \mathbf{j} + 3a^2z \mathbf{k}\). We need to evaluate the surface integral \(\int_{S} \mathbf{F} \cdot \mathbf{n} \ dS\) using the divergence theorem.

Step 2: Compute the Divergence of \(\mathbf{F}\)

The divergence theorem states that \(\int_{S} \mathbf{F} \cdot \mathbf{n} \ dS = \int_{V} abla \cdot \mathbf{F} \, dV\). Compute \( abla \cdot \mathbf{F} \) where \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(3a^2z) = 3x^2 + 3y^2 + 3a^2 \) over volume \(V\).

Step 3: Set up the Volume Integral

The volume \(V\) is within the cylinder \(x^2 + y^2 \leq a^2\) between \(z=0\) and \(z=1\). Transform the volume integral into cylindrical coordinates where \(x = r\cos \theta\), \(y = r\sin \theta\), and \(dV = r \, dr \, d\theta \, dz\). The limits for \(r\) are from 0 to \(a\), \(\theta\) from 0 to \(2\pi\), and \(z\) from 0 to 1.

Step 4: Evaluate the Volume Integral

Evaluate the integral: \[ \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{a} (3r^2 \cos^2 \theta + 3r^2 \sin^2 \theta + 3a^2) r \, dr \, d\theta \, dz. \]The expression \(3r^2 \cos^2 \theta + 3r^2 \sin^2 \theta\) simplifies to \(3r^2\) (since \(\cos^2 \theta + \sin^2 \theta = 1\)).This simplifies the integral to \[ \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{a} (3r^3 + 3a^2r) \, dr \, d\theta \, dz.\]

Step 5: Simplify and Solve the Integrals

Start with \(r\):\[ \int_{0}^{a} (3r^3 + 3a^2r) \, dr = \left[ \frac{3}{4}r^4 + \frac{3a^2}{2}r^2 \right]_{0}^{a} = \frac{3}{4}a^4 + \frac{3a^4}{2}.\]Evaluate \(\theta\):\[ \int_{0}^{2\pi} d\theta = 2\pi.\]Evaluate \(z\):\[ \int_{0}^{1} dz = 1.\]

Step 6: Combine and Calculate the Result

Multiplying these results, we have:\[ 1 \times 2\pi \times \left( \frac{3}{4}a^4 + \frac{3a^4}{2} \right) = 2\pi \left( \frac{3}{2}a^4 \right) = 3\pi a^4. \]The surface integral is thus \(3\pi a^4\).

Key Concepts

Surface Integral

A surface integral is a way to calculate the flow of a vector field across a surface. In this exercise, we are using the surface integral to find how much of the vector field \( \mathbf{F} \) passes through a given surface \( S \). The formula for a surface integral over a surface \( S \) with a vector field \( \mathbf{F} \) is given by:

• \( \int_{S} \mathbf{F} \cdot \mathbf{n} \, dS \)

Here, \( \mathbf{n} \) is the outward normal vector to the surface \( S \), and \( dS \) is a small area element of the surface. The dot product \( \mathbf{F} \cdot \mathbf{n} \) measures the component of \( \mathbf{F} \) that is perpendicular to the surface, indicating the direction and amount of flow through \( S \).

The exercise utilizes the Divergence Theorem to transform the challenging task of calculating the surface integral into a volume integral, which often simplifies the computation.

Cylinder

In this problem, the surface \( S \) is enclosed by a cylinder. The cylinder is a geometric shape with curved sides, usually described by a circular base. Here, the base is defined by the equation \( x^2 + y^2 = a^2 \), which signifies a circle with radius \( a \) in the \( xy \)-plane.

The cylinder's axis is aligned with the \( z \)-axis, meaning it extends vertically from \( z = 0 \) to \( z = 1 \). This gives the cylinder its height. The top and bottom faces of the cylinder are flat surfaces, making its boundary easy to define. Thus, the entire surface \( S \) is composed of three parts:

• The curved side \( x^2 + y^2 = a^2 \), \( 0 \leq z \leq 1 \)
• The top disk \( z = 1 \)
• The bottom disk \( z = 0 \)

Understanding the boundaries helps to set up integrals correctly when using cylindrical coordinates.

Cylindrical Coordinates

To simplify the computation of volume integrals, cylindrical coordinates are employed. This coordinate system is particularly useful for shapes like cylinders. Cylindrical coordinates \( (r, \theta, z) \) are defined as follows:

• \( r \) is the radial distance from the \( z \)-axis, corresponding to the radius in a circular motion.
• \( \theta \) is the angle measured in the \( xy \)-plane from the positive \( x \)-axis.
• \( z \) represents the height above the \( xy \)-plane.

The conversion from Cartesian to cylindrical coordinates can be expressed as:

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

Using these transformations, the volume element \( dV \) is expressed in cylindrical coordinates as \( r \, dr \, d\theta \, dz \). This makes it easier to evaluate integrals over regions bounded by cylinders like the one in this exercise.

Vector Field

A vector field assigns a vector to every point in space. In this exercise, the vector field \( \mathbf{F}(x, y, z) = x^3 \mathbf{i} + y^3 \mathbf{j} + 3a^2z \mathbf{k} \) describes the flow in a 3D space at each point \( (x, y, z) \). Each component of the vector field relates to a direction:

• \( x^3 \mathbf{i} \) represents the flow along the \( x \)-axis.
• \( y^3 \mathbf{j} \) signifies the flow along the \( y \)-axis.
• \( 3a^2z \mathbf{k} \) indicates the flow along the \( z \)-axis, dependent on the position \( z \) and constant \( a \).

The vector field provides data about the magnitude and direction of force at any given point in the space, which is crucial for evaluating how the field interacts with or passes through surfaces such as \( S \). By understanding the vector field, one can interpret how forces act spatially in physical systems.