Question

Compute the flux of \(\vec{F}\) across \(\mathcal{S}\). (If \(\mathcal{S}\) is not a closed surface, choose \(\vec{n}\) so that it has a positive z-component, unless otherwise indicated.) \(\mathcal{S}\) is the disk in the \(y-z\) plane with radius 1 , centered at (0,1,1) (choose \(\vec{n}\) such that it has a positive \(x\) -component); \(\overrightarrow{\vec{F}}=\langle y, z, x\rangle\)

Short answer

The flux is \( \pi \).

Step-by-step solution

Understand the Problem

We are asked to find the flux of the vector field \( \vec{F} = \langle y, z, x \rangle \) across the disk \( \mathcal{S} \) in the \( y-z \) plane with center at \( (0, 1, 1) \) and a radius of 1. The normal vector \( \vec{n} \) should have a positive \( x \)-component.

Define the Surface \( \mathcal{S} \)

The surface \( \mathcal{S} \) is a disk centered at \( (0,1,1) \) with a radius of 1 in the \( y-z \) plane. The equation for this disk is \( x = 0 \) with \( (y-1)^2 + (z-1)^2 \leq 1 \).

Determine the Normal Vector

Since the normal vector must have a positive \( x \)-component, we choose \( \vec{n} = \langle 1, 0, 0 \rangle \). This points out of the plane since the disk is perpendicular to the \( x \)-axis.

Set Up the Flux Integral

The flux of \( \vec{F} \) across \( \mathcal{S} \) is given by the integral \( \iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS \). Substituting \( \vec{F} = \langle y, z, x \rangle \) and \( \vec{n} = \langle 1, 0, 0 \rangle \), this becomes \( \iint_{\mathcal{S}} y \, dS \).

Parameterize the Surface \( \mathcal{S} \)

A point on the surface can be represented as \( (0, y, z) = (0, 1 + r \cos \theta, 1 + r \sin \theta) \) for \( r \) from 0 to 1 and \( \theta \) from 0 to \( 2\pi \).

Determine the Area Element \( dS \)

Since \( \mathcal{S} \) is in the \( y-z \) plane, the area element is \( dS = r \, dr \, d\theta \), corresponding to the radial coordinates.

Evaluate the Flux Integral

Substituting into the integral, \( \iint_{\mathcal{S}} (1 + r \cos \theta) \, r \, dr \, d\theta \), we compute: \[ \int_{0}^{2\pi} \int_{0}^{1} (1 + r \cos \theta) \, r \, dr \, d\theta = \int_{0}^{2\pi} \left[ \frac{r^2}{2} + \frac{r^3}{3} \cos \theta \right]_0^1 \, d\theta \]After evaluating, this reduces to: \[ \int_{0}^{2\pi} \left( \frac{1}{2} + \frac{1}{3} \cos \theta \right) \, d\theta \]. Solve this integral to find:\[ \left[ \frac{1}{2} \theta + \frac{1}{3} \sin \theta \right]_0^{2\pi} = \pi \].

Conclude

The computed flux of \( \vec{F} \) across \( \mathcal{S} \) with the specified normal direction is \( \pi \).

Key concepts

Vector Field

Vector fields are essential mathematical constructs used to represent a space where each point has a vector associated with it. These vectors can describe various quantities such as force, velocity, or any field of influence distributed across a region.

In our exercise, we encounter the vector field \( \vec{F} = \langle y, z, x \rangle \). Here, the vector assigned at each point in space has components that are functions of the coordinates \(y\), \(z\), and \(x\).

This particular vector field implies that at any location \((x, y, z)\), the vector's components are derived directly from these coordinates, creating a spatial map of directional flows. Understanding the structure of the vector field is crucial for applying operations like surface integrals, which rely on how these vectors interact with surfaces in their domain.

Surface Integral

Surface integrals are powerful tools in vector calculus used to calculate how a vector field interacts with a surface. Specifically, it integrates over a given surface, considering the field's values and the direction relative to the surface.

In our task, we calculate the flux of \( \vec{F} \) across the surface \( \mathcal{S} \). Flux inspects how much of the vector field "flows" through the surface. This is set up with the integral \( \iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS \), where \( \vec{n} \) is the normal vector to the surface. The core purpose of surface integrals is to condense information about the vector field's interaction with a surface into a single, insightful number, in this case, the computed flux which was found to be \( \pi \).

Such quantities are pivotal in physics and engineering, particularly in understanding phenomena like fluid flow, electromagnetic fields, and heat transfer.

Normal Vector

Normal vectors are perpendicular to a given surface, providing a direction that indicates outward or inward orientation relative to the surface. They play a vital role in defining how the field interacts with the surface during surface integrals.

For our exercise, the problem specifies that the surface \( \mathcal{S} \) must have a normal vector \( \vec{n} \) with a positive \( x \)-component. This decision ensures a specific directionality that impacts the calculation of flux. We chose \( \vec{n} = \langle 1, 0, 0 \rangle \), meaning that it points outward from the surface, aligned with the positive \( x \)-axis.

Understanding and choosing the correct normal vector is crucial because it ensures accurate calculation of the flux and reflects the real-world directionality of the vector field's impact on the surface.

Parametrization of Surface

Parametrization is the process of representing a surface through variables that can simplify calculations, typically with more intuitive geometric coordinates.

In our problem, the surface \( \mathcal{S} \) is a disk in the \( y-z \) plane centered at \((0, 1, 1)\). It's easier to describe this surface using polar coordinates \( (r, \theta) \) which correspond to the radius and angle, respectively.

Each point on the disk can be expressed as \((0, 1 + r \cos \theta, 1 + r \sin \theta)\) with \( r \) ranging from 0 to 1 and \( \theta \) from 0 to \( 2\pi \). The surface's area element \( dS \), crucial for the integral, converts to \( r \, dr \, d\theta \), reflecting how we compute areas in polar coordinates.

Understanding parametrization is essential as it transforms complex geometric shapes into calculable regions, making integral calculations feasible and more straightforward.