Problem 18 - Calculus Early Transcendentals

Question

Let \(\sigma\) be the closed surface consisting of the portion of the paraboloid \(z=x^{2}+y^{2}\) for which \(0 \leq z \leq 1\) and capped by the disk \(x^{2}+y^{2} \leq 1\) in the plane \(z=1 .\) Find the flux of the vector field \(\mathbf{F}(x, y, z)=z \mathbf{j}-y \mathbf{k}\) in the outward direction across \(\sigma .\)

Step-by-Step Solution

Answer

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

Step 1: Apply the Divergence Theorem

We will apply the divergence theorem to convert the surface integral into a volume integral. The divergence theorem states that \( \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{E} abla \cdot \mathbf{F} \, dV \), where \(E\) is the volume enclosed by \(\sigma\).

Step 2: Calculate the Divergence of F

Compute \( abla \cdot \mathbf{F} \) for the given vector field \( \mathbf{F}(x, y, z) = z \mathbf{j} - y \mathbf{k} \). The divergence \( abla \cdot \mathbf{F} \) is given by \( \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). Here, each component of \( \mathbf{F} \) does not depend on the variable of its partial derivative, so \( abla \cdot \mathbf{F} = 0 \).

Step 3: Apply the Divergence Theorem Result

Since \( abla \cdot \mathbf{F} = 0 \), the volume integral \( \iiint_{E} abla \cdot \mathbf{F} \, dV \) evaluates to zero. Hence, the flux of the vector field \( \mathbf{F} \) across the surface \( \sigma \) is also zero.

Step 4: Conclusion

By the divergence theorem, the flux of the vector field \( \mathbf{F} \) in the outward direction across the closed surface \( \sigma \) is zero. This occurs because the divergence of the vector field within the enclosed volume is zero.

Key Concepts

Flux

Flux is an important concept in vector calculus. It measures how much of a vector field passes through a given surface. Imagine standing in a river. The amount of water that flows past you every second can be thought of as the flux of the water's velocity field through a surface patch.

Mathematically, flux is determined by computing a surface integral. It involves taking the dot product of a vector field with a normal vector to the surface and then integrating over the surface area. For instance, if the vector field is \( \mathbf{F} \), and \( \sigma \) is the surface, then the flux through \( \sigma \) is given by the integral \(\ \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \\), where \( \mathbf{n} \) is the unit normal vector.

In the exercise provided, the flux of the vector field through the closed surface was calculated. The divergence theorem helped simplify this process by converting it to a volume integral. Converting surface integrals to volume integrals often makes calculations easier, especially when dealing with closed surfaces.

• Flux quantifies the "flow" of a vector field through a surface.
• It uses the surface integral of the dot product between the vector field and a normal vector.
• The Divergence Theorem can simplify the calculation of flux for closed surfaces.

Vector Field

A vector field represents a field that assigns a vector to each point in space. You can think of it like a breeze or fluid flow, where at each point in the field, a vector indicates the field's direction and magnitude.

For example, in this exercise, we dealt with the vector field \( \mathbf{F}(x, y, z) = z \mathbf{j} - y \mathbf{k} \). Each point in space has a vector that contributes to the flux calculation.

Understanding vector fields is essential in physics and engineering, as they can represent forces such as electromagnetic fields or gravitational fields. When analyzing vector fields, key features include:

• Direction: The direction of the vector at each point, showing where the vector is heading.
• Magnitude: The size or length of the vector, indicating the strength of the field at that point.
• Origin of Components: In the given field, \(z \mathbf{j} \) and \(-y \mathbf{k} \), the components don’t depend on x, showing the vector has constant x across the field.

These properties help in visualizing and calculating different quantities, like the flux across surfaces.

Surface Integral

A surface integral extends the concept of a line integral to two dimensions, integrating functions over a surface rather than along a path. It is fundamental for quantifying quantities like flux through a surface in a vector field.

To compute a surface integral of a vector field \(\mathbf{F} \), along a surface \(\sigma \), we utilize the formula \(\iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \\).

This integrates the dot product of \(\mathbf{F} \) with the unit normal vector \(\mathbf{n} \) over the surface area.

• Normal Vector (\(\mathbf{n}\)): A perpendicular vector to the surface, crucial for determining the orientation of the surface.
• Closed Surfaces: When the surface is closed, such as a sphere or closed paraboloid, surface integrals are particularly useful for applying the Divergence Theorem.

Understanding surface integrals helps in solving problems related to evaluating flow across different geometries and is essential in fields such as electromagnetism and fluid dynamics.

Volume Integral

Volume integrals allow you to integrate over three-dimensional regions. In the context of the Divergence Theorem, they convert surface integrals into simpler integrations over volumes. This is particularly helpful with zero or uniform divergence fields, as seen in the exercise where the divergence of the vector field was zero.

The conversion uses the divergence theorem: \[ \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{E} abla \cdot \mathbf{F} \, dV \]where \( E \) is the volume enclosed by the surface \( \sigma \). This relation stems from the idea that the total "outflow" from a volume is equal to the sum of all the sources (and sinks) within that volume.

Some points to remember about volume integrals:

• They integrate a scalar function over a three-dimensional space.
• Particularly useful when applying the Divergence Theorem to simplify flux calculations.
• In our exercise, it was used to demonstrate that the outward flux is zero due to zero divergence within the volume.

Volume integrals thus connect powerful mathematical concepts and physical interpretations, often simplifying complex calculations in physics and engineering.