Question

Rain on roofs Let \(z=s(x, y)\) define a surface over a region \(R\) in the \(x y\) -plane, where \(z \geq 0\) on \(R\). Show that the downward flux of the vertical vector field \(\mathbf{F}=\langle 0,0,-1\rangle\) across \(S\) equals the area of \(R .\) Interpret the result physically.

Short answer

Based on the solution provided, show that the downward flux of the vector field F across the surface S equals the area of the region R. Answer: Following the given step-by-step solution, we found that the downward flux of the vector field F = ⟨0, 0, -1⟩ across the surface S parameterized by r(x, y) = ⟨x, y, s(x, y)⟩ is equal to the area of the region R in the xy-plane. We computed the flux as the double integral of F · dS over R and found that it equals negative times the area of R. Physically, this result represents the amount of rain passing through the surface S and being collected in the region R, with the negative sign indicating the downward direction of the rain.

Step-by-step solution

Parameterize the surface

Let's parameterize the surface \(S\). Since it's defined over the region \(R\) in the xy-plane, we can use the following parameterization: $$ \mathbf{r}(x, y) = \langle x, y, s(x, y) \rangle, $$ where \((x, y) \in R\).

Find the normal vector

Now let's find the normal vector \(d\mathbf{S}\) on the surface \(S\). We can obtain this vector by finding the cross product of the partial derivatives of the parameterization function \(\mathbf{r}(x, y)\) with respect to \(x\) and \(y\), as follows: $$ d\mathbf{S} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y}. $$ First, let's compute the partial derivatives: $$ \frac{\partial \mathbf{r}}{\partial x} = \langle 1, 0, \frac{\partial s}{\partial x} \rangle, \\ \frac{\partial \mathbf{r}}{\partial y} = \langle 0, 1, \frac{\partial s}{\partial y} \rangle. $$ Now, we can compute the cross product: $$ d\mathbf{S} = \langle 1, 0, \frac{\partial s}{\partial x} \rangle \times \langle 0, 1, \frac{\partial s}{\partial y} \rangle = \left\langle -\frac{\partial s}{\partial x}, -\frac{\partial s}{\partial y}, 1 \right\rangle. $$

Compute the flux

Now we can compute the flux of the vector field \(\mathbf{F}\) across the surface \(S\) as follows: $$ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_R \langle 0, 0, -1 \rangle \cdot \left\langle -\frac{\partial s}{\partial x}, -\frac{\partial s}{\partial y}, 1 \right\rangle \, dx \, dy. $$ Calculate the dot product: $$ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_R (-1)(1) \, dx \, dy = -\iint_R dx \, dy. $$

Find the area of R

The area of the region \(R\) can be calculated as the following double integral: $$ \text{Area}(R) = \iint_R dx \, dy. $$

Compare the flux and the area of R

Comparing the flux and the area of \(R\), we can see that: $$ \text{Flux} = -\text{Area}(R), $$ which implies that the downward flux of the vector field \(\mathbf{F}\) across the surface \(S\) equals the area of the region \(R\).

Physical interpretation

The result can be physically interpreted as follows: The downward flux of the vertical vector field \(\mathbf{F} = \langle 0, 0, -1 \rangle\) across the surface \(S\) represents the amount of rain passing through the surface \(S\). This amount is equal to the area of the region \(R\), which represents the total "catchment area" that collects the rain in the xy-plane. The negative sign indicates that the rain is falling in the downward direction, following the vector field \(\mathbf{F}\).

Key concepts

Vector Fields

A vector field is a function that assigns a vector to each point in space. In this exercise, we are dealing with a simple vector field, \(\mathbf{F} = \langle 0, 0, -1 \rangle\). This particular vector field represents a uniform downward flow, such as gravity or rainfall. Because the only non-zero component is the \(-1\) in the z-direction, it implies a consistent downward vector at every point in the space defined by the surface \(S\).

Vector fields are key in understanding how vectors change across a given region or surface, and they are especially useful in physics and engineering for analyzing forces like electric and magnetic fields, or fluid flow.

In this problem, the downward vector field \(\mathbf{F}\) models rain falling straight down on a surface. The exercise demonstrates how to compute the flux of this vector field through a defined surface by first understanding the nature of the vector field involved.

Flux Calculation

Flux quantifies how much of a vector field passes through a surface. In simple terms, it measures the flow of the vector field across a surface. In our exercise, the goal is to compute the downward flux of the vector field \( \mathbf{F} = \langle 0, 0, -1 \rangle \) across the surface \(S\).

The flux is calculated using the surface integral
\[ \iint_S \mathbf{F} \cdot d\mathbf{S} \]
This integral evaluates the dot product between the vector field and a normal vector to the surface \(d\mathbf{S}\). The dot product essentially gives us the component of the vector field that is perpendicular to the surface. By integrating this over the entire surface, we determine the total flux passing through.

In the solution, it is shown that the downward flux equals the negative of the area of the region \(R\). This is because the vector field \(\mathbf{F}\) is directed downward, and the orientation of our normal vector \(d\mathbf{S}\) results in a negative sign.

Parameterization

Parameterization is the process of describing a surface using a set of parameters. In our exercise, the surface \(S\) is defined over a region \(R\) in the \(xy\)-plane using a parametric representation.

This specific representation is given by the vector function
\( \mathbf{r}(x, y) = \langle x, y, s(x, y) \rangle \)
Here, \(x\) and \(y\) vary within the region \(R\), and \(s(x,y)\) defines the height of the surface above each point \((x, y)\) in \(R\).

Parameterization not only describes the surface, but it also facilitates the computation of other important quantities such as derivatives, normal vectors, and surface integrals. By using these parameters, complex shapes and surfaces can be analyzed in a manageable way, enabling easier calculation of flux and other attributes.

Normal Vector

Finding the normal vector is crucial for calculating flux because it provides the necessary orientation for the dot product in the surface integral. In this solution, the normal vector \(d\mathbf{S}\) on the surface \(S\) is derived from the cross product of the partial derivatives of the parameterization function \(\mathbf{r}(x, y)\).

The partial derivatives are:
\[ \frac{\partial \mathbf{r}}{\partial x} = \langle 1, 0, \frac{\partial s}{\partial x} \rangle \]
\[ \frac{\partial \mathbf{r}}{\partial y} = \langle 0, 1, \frac{\partial s}{\partial y} \rangle \]
The normal vector is then computed as their cross product:
\[ d\mathbf{S} = \langle -\frac{\partial s}{\partial x}, -\frac{\partial s}{\partial y}, 1 \rangle \]
This vector is perpendicular to the surface at each point, which is essential for determining how the vector field interacts with the surface.

The direction and magnitude of this normal vector also influence the sign and value of the flux integral, giving insight into the net flow direction and strength through the surface.