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 square in space with corners at \((0,0,0),(1,0,0),\) (1,0,1) and (0,0,1) (choose \(\vec{n}\) such that it has a positive \(y\) -component \() ; \vec{F}=\langle 0,-z, y\rangle .\)

Short answer

The flux of \( \vec{F} \) across \( \mathcal{S} \) is \( -\frac{1}{2}. \)

Step-by-step solution

Understanding the Problem

We need to calculate the flux of the vector field \( \vec{F} = \langle 0, -z, y \rangle \) across the square plane \( \mathcal{S} \) with corners at \( (0,0,0), (1,0,0), (1,0,1), (0,0,1) \). The plane is parallel to the \( xz \)-plane and is vertical with a normal vector \( \vec{n} \) having a positive \( y \)-component.

Determine Surface Normal

The square \( \mathcal{S} \) lies in the \( xz \)-plane and is vertical. A vector normal to a surface parallel to the \( xz \)-plane that's vertical and extends in the \( y \)-direction would be \( \vec{n} = \langle 0, 1, 0 \rangle \). This is chosen to ensure the normal has a positive \( y \)-component.

Set Up the Flux Integral

The flux of \( \vec{F} \) through \( \mathcal{S} \) is given by \( \iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS \). Substitute for \( \vec{F} = \langle 0, -z, y \rangle \) and \( \vec{n} = \langle 0, 1, 0 \rangle \). The dot product is \( \vec{F} \cdot \vec{n} = (-z)(1) + (y)(0) = -z \). The integral thus becomes \( \iint_{\mathcal{S}} -z \, dS \).

Parameterize Surface and Evaluate Integral

Parameterize the surface \( \mathcal{S} \) by choosing \( x \) and \( z \) as the parameters: \( \vec{r}(x, z) = \langle x, 0, z \rangle \), where \( 0 \leq x \leq 1 \) and \( 0 \leq z \leq 1 \). The differential area element \( dS \) on this parameterized surface is \( dx \, dz \). Evaluating the integral: \[ \iint_{0}^{1}\iint_{0}^{1} -z \, dx \, dz = \int_{0}^{1} \left( \int_{0}^{1} -z \, dx \right) dz. \] The inner integral evaluates to \( \int_{0}^{1} -z \, dx = -z \).

Integrate Outer Integral

Now integrate the result from the inner integral with respect to \( z \): \[ \int_{0}^{1} -z \, dz = \left[ -\frac{z^2}{2} \right]_{0}^{1} = -\frac{1}{2}. \] This gives the flux of \( \vec{F} \) across \( \mathcal{S} \).

Key concepts

Vector Field

A vector field is a function that assigns a vector to every point in space. In the context of this exercise, we are given a vector field \( \vec{F} = \langle 0, -z, y \rangle \). This means that at any point \((x, y, z)\), the vector field defines a vector pointing in a specific direction, influenced by the coordinates of the point. Here, the vector has no \(x\)-component, a negative \(z\)-component (pointing downwards when \(z > 0\)), and a positive \(y\)-component (pointing in the upwards direction when \(y > 0\)).
Understanding the behavior and directionality of vector fields is crucial, especially when evaluating their interactions, such as their flux through a surface.

Flux Integral

A flux integral computes the flow of a vector field through a specific surface. The concept involves integrating the dot product of the vector field \( \vec{F} \) and the surface's normal vector \( \vec{n} \) over the area of the surface. In simpler terms, it measures how much of the vector field passes through the surface.
For the given problem, we need to evaluate the flux integral \( \iint_{\mathcal{S}} \vec{F} \cdot \vec{n} \, dS \). This step requires determining the part of the vector field that is perpendicular to the surface at each point, through the dot product. The importance of the flux integral lies in its ability to capture and quantify this notion of flow across the surface area.

Surface Parameterization

Parameterization is a way to represent a surface using parameters that trace out the surface within some bounds. In this exercise, we use \(x\) and \(z\) as parameters to describe the surface \(\mathcal{S}\). The parameterized form is \( \vec{r}(x, z) = \langle x, 0, z \rangle \), which effectively places the surface in the \(xz\)-plane from \(x = 0\) to \(x = 1\) and \(z = 0\) to \(z = 1\).
Parameterization simplifies the process of performing integrals over surfaces, as it allows the expression of a multi-dimensional surface in terms of lower-dimensional parameters. It also helps establish the limits of integration and find expressions for the differential area elements \(dS\). Here, \(dS = dx \, dz\) corresponds to small patches of the surface being integrated over.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This single number, or scalar, represents the magnitude of one vector projected onto another. For vectors \( \vec{a} \) and \( \vec{b} \), the dot product is \( \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
In the exercise, the dot product \( \vec{F} \cdot \vec{n} \) simplifies the flux integral by reducing the computation to a scalar value. This value reflects how much of the vector field \( \vec{F} \) penetrates through the surface at a particular point. By breaking \(\vec{F}\) and \(\vec{n}\) into components, calculating the dot product allows us to determine the effective component of the vector field that contributes to the flux through the surface.