Problem 24 - Calculus Early Transcendentals

Question

Find the flux of \(\mathbf{F}\) across the surface \(\sigma\) by expressing \(\sigma\) parametrically. \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} ; \sigma\) is the portion of the cylinder \(x^{2}+z^{2}=1\) between the planes \(y=1\) and \(y=-2,\) oriented by outward unit normals.

Step-by-Step Solution

Answer

The flux of \( \mathbf{F} \) across \( \sigma \) is \( 6\pi \).

Step 1: Parametrize the Surface

The surface \( \sigma \) is a part of a cylinder defined by the equation \( x^2 + z^2 = 1 \), with \( y \) varying from \(-2\) to \( 1\). It can be parametrized using cylindrical coordinates. Let:\[x = \cos\theta, \quad z = \sin\theta, \quad y = t\]where \( \theta \) is the parameter for the circular section of the cylinder, ranging from \( 0 \) to \( 2\pi \), and \( t \) is the parameter along the height of the cylinder, ranging from \( -2 \) to \( 1 \).

Step 2: Calculate the Partial Derivatives

Compute the partial derivatives of the parametrization with respect to \( \theta \) and \( t \):\[\frac{\partial \sigma}{\partial \theta} = \left(-\sin\theta, 0, \cos\theta\right)\]\[\frac{\partial \sigma}{\partial t} = (0, 1, 0)\]These partial derivatives form two tangent vectors to the surface.

Step 3: Compute the Normal Vector

The normal vector to the surface can be found by taking the cross product of the two tangent vectors:\[\left(\frac{\partial \sigma}{\partial \theta}\right) \times \left(\frac{\partial \sigma}{\partial t}\right) = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-\sin\theta & 0 & \cos\theta \0 & 1 & 0\end{vmatrix} = (-\cos\theta, 0, -\sin\theta)\]This is the outward normal vector since it points away from the center of the cylinder.

Step 4: Set Up the Flux Integral

The surface integral for flux is:\[\iint_\sigma \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F} \cdot \mathbf{n} \, dA\]where \( \mathbf{n} \) is the unit normal vector. Here \( \mathbf{F}(\sigma(\theta, t)) = (\cos\theta, t, \sin\theta) \). Therefore, the dot product is:\[\mathbf{F} \cdot \mathbf{n} = (\cos\theta, t, \sin\theta) \cdot (-\cos\theta, 0, -\sin\theta) = -\cos^2\theta - \sin^2\theta = -1\]This simplifies the integral significantly.

Step 5: Evaluate the Integral

The double integral becomes:\[-\iint_D dA\] where \( D \) is the rectangular domain \( 0 \leq \theta \leq 2\pi \) and \( -2 \leq t \leq 1 \). The area \( dA = d\theta dt \), so:\[\int_{0}^{2\pi}\int_{-2}^{1} (-1) \, dt \, d\theta = \int_{0}^{2\pi}\left[-t\right]_{-2}^{1} d\theta = \int_{0}^{2\pi} (1 - (-2)) \, d\theta = 3\int_{0}^{2\pi} d\theta = 3\times 2\pi = 6\pi \]

Key Concepts

Parametric Surfaces

A parametric surface is a representation of a two-dimensional surface in three-dimensional space using parameters (usually two). It is a powerful method for defining complex shapes like cylinders, spheres, and more, using equations that evolve in terms of these parameters. For the given problem, the surface \( \sigma \) is part of a cylinder characterized by the equation \( x^2 + z^2 = 1 \). We can imagine this as being traced by shifting a circle (the cross-section) up and down along the \( y \)-axis.

To parametrize this specific surface, we use cylindrical coordinates because they suit cylindrical shapes seamlessly. Cylindrical coordinates let us represent points on the circular base easily using the angle \( \theta \) and the height \( t \). Here's the breakdown:

• \( x = \cos\theta \): This represents the \( x \) component, tracing a circle in the \( xz \)-plane.
• \( z = \sin\theta \): This is the \( z \) component, also part of the circular trace.
• \( y = t \): This component reflects the vertical shift along the height of the cylinder.

The angle \( \theta \) varies from \( 0 \) to \( 2\pi \) (which completes a full circle), while the height \( t \) spans from \( -2 \) to \( 1 \) along the \( y \)-axis. This gives us a complete parametric representation for the cylindrical portion \( \sigma \).

Cylindrical Coordinates

Cylindrical coordinates are an extension of polar coordinates, useful for problems involving shapes like cylinders or spirals. This system incorporates an angle \( \theta \), a radial distance \( r \), and a vertical component \( z \) or \( t \) instead of \( x \), \( y \), \( z \) used in Cartesian coordinates. This is especially handy for surfaces of revolution, like our cylinder.

Understanding cylindrical coordinates involves three parts:

• The radial distance \( r \) is often the square root of the sum of squares of the Cartesian components \( x \) and \( z \), signifying the distance from the central axis (in this case, the \( y \)-axis).
• The angle \( \theta \) is measured from a reference direction, often the positive \( x \)-axis, counterclockwise in the \( xz \)-plane.
• The height or axial component \( y = t \) represents position along the axis of the cylinder, perpendicular to the radial directions.

In cylindrical problems, it's common to replace \( x \) and \( z \) components with \( r\cos\theta \) and \( r\sin\theta \), respectively. Here, for the unit circle, \( r = 1 \), so \( x = \cos\theta \) and \( z = \sin\theta \). These formulas simplify the process of parametric representation and calculation.

Surface Integrals

Surface integrals extend the concept of line integrals to surfaces, allowing the calculation of flux through a surface. In physical terms, flux measures how much of a vector field passes through a given surface. This technique is crucial in fields like electromagnetism and fluid dynamics to quantify, for example, the flow of a fluid across a surface.

For a surface defined parametrically, we first derive tangent vectors from partial derivatives of the parametric equations. The cross product of these tangent vectors gives the normal vector necessary for surface integrals.

In this problem, the normal vector is derived by cross-multiplying:
\[ \left( \frac{\partial \sigma}{\partial \theta} \right) \times \left( \frac{\partial \sigma}{\partial t} \right) \]
resulting in \( (-\cos\theta, 0, -\sin\theta) \), pointing outward, away from the cylinder's center. This aligns with the orientation requirement of outward unit normals.

With the flux formula \( \iint_\sigma \mathbf{F} \cdot d\mathbf{S} \), the dot product simplifies the integral. We find:
\[ \mathbf{F} \cdot \mathbf{n} = -\cos^2\theta - \sin^2\theta = -1 \]
This constant simplifies the calculation considerably, leading directly to the final result of the flux integral. The result of \( 6\pi \) emerges from integrating over the defined parameter space \( 0 \leq \theta \leq 2\pi \) and \( -2 \leq t \leq 1 \).