Problem 11 - Calculus Early Transcendentals

Question

Use the Divergence Theorem to find the flux of F across the surface ? with outward orientation. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=(x-z) \mathbf{i}+(y-x) \mathbf{j}+(z-y) \mathbf{k} ; \sigma \text { is the sur- }} \\ {\text { face of the cylindrical solid bounded by } x^{2}+y^{2}=a^{2}} \\ {z=0, \text { and } z=1}\end{array} $$

Step-by-Step Solution

Answer

The flux is \(3\pi a^2\).

Step 1: Understand the Divergence Theorem

The Divergence Theorem states that the flux of a vector field \( \mathbf{F} \) across a closed surface \( \sigma \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( \sigma \). The formula is: \[ \iint_{\sigma} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} abla \cdot \mathbf{F} \, dV \]. Here, \( \mathbf{F}(x, y, z) = (x-z) \mathbf{i} + (y-x) \mathbf{j} + (z-y) \mathbf{k} \).

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

The divergence is obtained by taking the partial derivative with respect to each component: \[ abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x - z) + \frac{\partial}{\partial y}(y - x) + \frac{\partial}{\partial z}(z - y) \]. Calculate each term individually: for \( x - z \), the partial derivative is 1; for \( y - x \), the partial derivative is 1; and for \( z - y \), the partial derivative is 1. Thus, \( abla \cdot \mathbf{F} = 1 + 1 + 1 = 3 \).

Step 3: Set Up the Bounds for the Triple Integral

The cylindrical volume is defined by \( x^2 + y^2 \leq a^2 \), \( 0 \leq z \leq 1 \). In cylindrical coordinates: \( x = r\cos\theta \), \( y = r\sin\theta \), \( z = z \), with bounds \( 0 \leq r \leq a \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq 1 \).

Step 4: Evaluate the Triple Integral

Convert the integral into cylindrical coordinates: \[ \iiint_V abla \cdot \mathbf{F} \, dV = \iiint_V 3 \, dV = 3 \int_{0}^{2\pi} \int_{0}^{a} \int_{0}^{1} r \, dz \, dr \, d\theta \]. First integrate with respect to \( z \): \( \int_{0}^{1} 1 \, dz = 1 \). Substituting back: \[ 3 \int_{0}^{2\pi} \int_{0}^{a} r \, dr \, d\theta \]. Now integrate with respect to \( r \): \( \int_{0}^{a} r \, dr = \frac{a^2}{2} \). Substituting back: \[ 3 \int_{0}^{2\pi} \frac{a^2}{2} \, d\theta \]. Integrate with respect to \( \theta \): \( \int_{0}^{2\pi} 1 \, d\theta = 2\pi \).

Step 5: Calculate the Final Result

Completing the integration, we have \[ 3 \times \frac{a^2}{2} \times 2\pi = 3\pi a^2 \]. This is the flux of the vector field across the surface \( \sigma \).

Key Concepts

Flux Calculation

The Divergence Theorem is a powerful tool in vector calculus used to calculate the flux of a vector field across a closed surface. Flux essentially measures the quantity of a vector field passing through a surface. When addressing problems related to flux calculation, we frequently utilize the Divergence Theorem. This theorem simplifies surface integrals into volume integrals, making computations much easier for closed surfaces.

When working with a vector field \( \mathbf{F} \), we first calculate its divergence, \( abla \cdot \mathbf{F} \). This value represents how much the vector field spreads out from a point. Once the divergence is computed, we apply it to a volume integral over the area enclosed by the surface. This simplifies the complex surface integral into a more manageable form. In our case, the divergence of \( \mathbf{F} \) was determined to be \( 3 \), so the flux integral became \( \iiint_V 3 \, dV \).

• Flux measures vector field intensity through a surface.
• Divergence Theorem simplifies surface integrals to volume integrals.
• Calculation requires finding the divergence \( abla \cdot \mathbf{F} \).

By understanding these key points, calculating the flux for vector fields over complex surfaces becomes approachable and more understandable.

Cylindrical Coordinates

Cylindrical coordinates are a system used to simplify problems with symmetries about an axis, typically the z-axis. This system uses a combination of linear and angular coordinates, expressed as \( (r, \theta, z) \), which correspond to the radial distance, the angular position in the xy-plane, and the height along the z-axis, respectively.

The conversion from Cartesian to cylindrical coordinates involves:

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

In exercises like these, switch to cylindrical coordinates when dealing with surfaces such as cylinders or circular regions. It significantly simplifies volume integrations, as was necessary for the cylindrical solid in the problem. Once in cylindrical coordinates, the bounds for \( r \) are set by the radius of the cylinder \( 0 \leq r \leq a \), for \( \theta \), the full rotation \( 0 \leq \theta \leq 2\pi \), and for \( z \) by its height \( 0 \leq z \leq 1 \).

• Converts Cartesian (x, y, z) to cylindrical (r, \( \theta \), z).
• Simplifies integration for symmetric cylindrical volumes.

Vector Field

A vector field, denoted \( \mathbf{F}(x, y, z) \), assigns a vector to every point in space. In the given exercise, \( \mathbf{F} = (x - z) \mathbf{i} + (y - x) \mathbf{j} + (z - y) \mathbf{k} \). Such fields often arise in physics to describe forces or velocities in a space. Understanding the direction and magnitude at different points is crucial to solving problems related to vector fields.

In analyzing the vector field for the problem, it involved computing the divergence \( abla \cdot \mathbf{F} \). This measures how much the field converges or diverges from a point, offering insights into its behavior across the surface.

• Vector fields map vectors to space points.
• Useful in representing physical phenomena like fluid flow and electromagnetic fields.
• Divergence \( abla \cdot \mathbf{F} \) provides information on field behavior.

With vector fields, understanding their mathematical structure helps in applications such as flux calculations and various interdisciplinary fields.