Problem 7 - Calculus Early Transcendentals

Question

Find the flux of the vector field \(\mathbf{F}\) across \(\sigma\) \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+2 z \mathbf{k} ; \sigma\) is the portion of the surface \(z=1-x^{2}-y^{2}\) above the \(x y-\) plane, oriented by upward normals.

Step-by-Step Solution

Answer

The flux of \( \mathbf{F} \) across \( \sigma \) is \( 2\pi \).

Step 1: Understand the Problem

We need to find the flux of the vector field \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + 2z \mathbf{k} \) across the surface \( \sigma \), which is a portion of the paraboloid \( z = 1 - x^2 - y^2 \) above the xy-plane.

Step 2: Setup the Surface Integral

The flux across the surface \( \sigma \) is given by the surface integral: \( \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the outward normal vector to the surface.

Step 3: Parametrize the Surface

We express the surface using the parameterization \( \mathbf{r}(x, y) = \langle x, y, 1 - x^2 - y^2 \rangle \). \( x \) and \( y \) vary such that \( x^2 + y^2 \leq 1 \), which forms a disk in the xy-plane.

Step 4: Calculate the Normal Vector

The upward unit normal vector \( \mathbf{n} \) to the surface is calculated using the partial derivatives: \( \mathbf{r}_x \times \mathbf{r}_y = \langle 2x, 2y, 1 \rangle \). Normalize this to get: \( \mathbf{n} = \langle \frac{2x}{\sqrt{4x^2 + 4y^2 + 1}}, \frac{2y}{\sqrt{4x^2 + 4y^2 + 1}}, \frac{1}{\sqrt{4x^2 + 4y^2 + 1}} \rangle \).

Step 5: Compute \( \mathbf{F} \cdot \mathbf{n} \)

Substitute \( \mathbf{F} = \langle x, y, 2(1-x^2-y^2) \rangle \) and \( \mathbf{n} = \langle 2x, 2y, 1 \rangle/\sqrt{4x^2 + 4y^2 + 1} \), then simplify the dot product \( \mathbf{F} \cdot \mathbf{n} \).

Step 6: Evaluate the Integral

Compute the double integral: \( \iint_{D} (4x^2 + 4y^2 + 2 - 6x^2 - 6y^2) \, dydx \), adjusting for \( dS \) in polar coordinates, where \( D \) is the unit disk \( x^2 + y^2 \leq 1 \).

Step 7: Change to Polar Coordinates

In polar coordinates, where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), the integral becomes \( \int_0^{2\pi} \int_0^1 (-2r^2 + 2) \cdot r \, drd\theta \).

Step 8: Integrate

Evaluate \( \int_0^{2\pi} \int_0^1 (-2r^3 + 2r) \, drd\theta \), integrating first with respect to \( r \) and then with respect to \( \theta \).

Key Concepts

Vector Field

A vector field is an assignment of a vector to each point in a subset of space. In three-dimensional space, a common representation for a vector field is \( \mathbf{F}(x, y, z) = P(x, y, z) \mathbf{i} + Q(x, y, z) \mathbf{j} + R(x, y, z) \mathbf{k} \). Here, - \( P, Q, \) and \( R \) are functions of \( x, y, \) and \( z \), indicating the components of the vector field.- \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors in the x, y, and z directions, respectively.Utilizing this framework, each point in space has a vector associated with it, determining its magnitude and direction. This is crucial in physics and engineering, as it helps describe phenomena like fluid flow, electromagnetic fields, and gravitational forces.
In our original problem, the vector field provided is \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + 2z \mathbf{k} \). This specific vector field can represent many real-world scenarios, such as wind or water current where the strength and direction change depending on the spatial location.
Understanding the vector field allows us to commence with the essential task of calculating how these vectors "flow" across a surface, which brings us to surface integrals.

Surface Integral

Surface integrals extend the idea of an integral to compute the sum of a field over a curved surface in three-dimensional space. Essentially, it measures the cumulative effect of the field across a surface. Visually, you can compare it to covering a surface with a net of vectors and summing up their influence.
To calculate the flux of a vector field across a surface, we use the surface integral. The formula for this is:\[ \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \] where:

• \( \mathbf{n} \) is the unit normal vector to the surface, pointing outward, ensuring the direction of the surface is accounted for.
• \( dS \) is the differential area of the surface element.

This operation gives the total flow (or flux) passing through the surface \( \sigma \), such as how much air passes through a net. In mathematical terms, the integral encompasses the dot product of the vector field with the surface's normal vector, scaling the vectors according to their perpendicularity to the surface.
For example, if the vector field is "flowing" parallel to the surface, it contributes nothing to the flux. Conversely, if it's perpendicular, its entire magnitude counts.
In the exercise, this concept is harnessed to find the integral over the surface defined by \( z = 1 - x^2 - y^2 \), above the xy-plane.

Parametrization

Parametrization is a technique that expresses a surface or curve using parameters, typically to simplify complex integrals. It transforms a two-dimensional surface in space into a manageable formula.
In our context, we need to portray the surface \( z = 1 - x^2 - y^2 \) in terms of parameters \( x \) and \( y \). This is given by the parameterization:\[ \mathbf{r}(x, y) = \langle x, y, 1 - x^2 - y^2 \rangle \]Within this framework:

• \( x \) and \( y \) act as parameters that vary across a region in the plane, here \( x^2 + y^2 \leq 1 \).
• The function \( 1 - x^2 - y^2 \) governs the vertical position (the height) of points on the surface.

This structure hinges on writing the surface based on simpler, well-behaved variables. It aids in defining the surface's geometry, letting us compute derivatives to find the tangent plane or normal vector, further employed in our surface integral calculations.
To get a whole picture, we need the partial derivatives of \( \mathbf{r}(x, y) \), indicating changes along our parameters. These derivatives are used subsequently to find the cross product, providing the normal vector to the surface.

Polar Coordinates

Polar coordinates are a system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This is particularly useful for circular or radial symmetries, which are harder to manage using Cartesian coordinates \( (x, y) \).
The transformation from Cartesian (x, y) to polar coordinates \((r, \theta)\) is given by:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

Here, \( r \) represents the radial distance from the origin, while \( \theta \) specifies the angle from the positive x-axis. This conversion makes integration over circular domains straightforward.
In the exercise, polar coordinates help evaluate the integral over the circular disk \( x^2 + y^2 \leq 1 \). By converting to polar, the circular limits of integration turn into constants, simplifying further calculations. Polar coordinates are quintessential in handling surfaces like paraboloids, as they naturally complement the symmetry in the problem setup.