Problem 19 - Thomas Calculus

Question

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. Parabolic cylinder \(\mathbf{F}=z^{2} \mathbf{i}+x \mathbf{j}-3 z\) outward (normal away from the \(x\) -axis) through the surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=1,\) and \(z=0\)

Step-by-Step Solution

Answer

The flux across the surface is 48 units.

Step 1: Identify the Surface and Boundaries

The surface given is a parabolic cylinder described by the equation \(z = 4 - y^2\), bounded by the planes \(x = 0\), \(x = 1\), and \(z = 0\). In the \(x\)-direction, the cylinder is between \(x = 0\) and \(x = 1\), and in the \(z\)-direction, it is capped at \(z = 0\).

Step 2: Parametrize the Surface

To parametrize the surface, use variables \(u\) and \(v\) as follows: \(x = u\), \(y = v\), and \(z = 4 - v^2\), where \(u\) ranges from \(0\) to \(1\) and \(v\) satisfies \(0 \leq v^2 \leq 4\), i.e., \(-2 \leq v \leq 2\).

Step 3: Calculate the Normal Vector

The surface is parametrize by \(\mathbf{r}(u, v) = \langle u, v, 4 - v^2 \rangle\). Compute the tangent vectors: \(\mathbf{r}_u = \langle 1, 0, 0 \rangle\) and \(\mathbf{r}_v = \langle 0, 1, -2v \rangle\). The cross product, \(\mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v = \langle 0, 2v, -1 \rangle\), gives the normal vector.

Step 4: Calculate the Flux Integral

The vector field is \(\mathbf{F} = z^2 \mathbf{i} + x \mathbf{j} - 3z\). The dot product needed is \(\mathbf{F} \cdot \mathbf{n} = (4 - v^2)^2 \cdot 0 + u \cdot 2v - 3(4 - v^2)\cdot(-1)\). Evaluate the double integral: \[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma = \int_{u=0}^1\int_{v=-2}^{2} \Big(2uv + 3(4 - v^2)\Big) \, dv \, du \]

Step 5: Integrate Over the Region

First, solve the inner integral in terms of \(v\): \[ \int_{v=-2}^{2} \Big(2uv + 12 - 3v^2\Big) \, dv = 12v + uv^2 - v^3 \Big|_{-2}^{2} \]Evaluating from \(v = -2\) to \(v = 2\), and simplifying, calculate the integration. Then compute the outer integral for \(u\) from \(0\) to \(1\).

Step 6: Evaluate and Simplify

Solving the definite integrals for both \(v\) and \(u\) gives the required flux across the surface. Simplify the expressions at each stage to find the solution. The calculations should reveal that the flux through the specified surface comes out to a single value.

Key Concepts

Parametrization of Surfaces

In calculus, particularly in the realm of multivariable calculus, surfaces can be challenging to handle due to their varying shapes and orientations. Parametrization is a tool that allows us to describe these surfaces in a more functional way. Parametrization involves expressing the coordinates of points on a surface using one or more parameters. For the exercise at hand, we use \(u\) and \(v\) as parameters to express the positions of points on the parabolic cylinder surface.

• This is done by setting \(x = u\), \(y = v\), and \(z = 4-v^2\).
• The parameters \(u\) and \(v\) determine every possible point on the curved surface within the given boundaries.

Understanding this method helps simplify complex surface calculations, such as finding flux integrals, by reducing them to calculations over a plane-like coordinate system. This is crucial as it makes integration over the surfaces manageable by converting a three-dimensional problem into a more familiar two-dimensional one.

Vector Fields

A vector field is essentially a function that assigns a vector to every point in space. In this exercise, the vector field given is \(\mathbf{F} = z^2 \mathbf{i} + x \mathbf{j} - 3z\). Each vector has components dependent on the coordinates of the point at which it's being considered.

• Vector fields are vital in physics and engineering as they represent quantities like velocity fields of fluids, magnetic fields, or force fields.
• They help us understand how, for instance, a fluid moves through space at any given point.

Working with these fields involves tasks like computing flux, which is the amount of vector field passing through a surface. In other words, flux quantifies the flow through a surface in the context of the defined vector field. For the exercise's purpose, understanding the formation and interaction of vector fields with surfaces will lead to proper calculations of the flux across the described parabolic cylinder.

Parabolic Cylinders

A parabolic cylinder is a surface that is shaped like a parabola extended in one direction. In more clear terms, you can think of it as a parabola that stretches infinitely along a third axis. The equation \(z = 4 - y^2\) describes a parabolic curve in the yz-plane, which continues along the x-axis, creating a cylinder-like shape.

• Parabolic cylinders occur in various physical systems, like in optics, where mirrors can be parabolic to focus light.
• They possess a simple structure when compared with some other 3D surfaces, making them easier to parametrize using basic variables.

Understanding parabolic cylinders is essential because it helps in imagining how the surface is shaped and aids in visualizing integration over these surfaces. In this exercise, knowing the boundaries of the parabolic cylinder (like planes \(x = 0\), \(x = 1\), and \(z = 0\)) helps to establish limits for integration, significantly simplifying the evaluation of integrals for flux.

Normal Vectors

To calculate flux, normal vectors are incredibly important since they represent the orientation of a surface at a given point. A normal vector \(\mathbf{n}\) to a surface is a vector perpendicular to the surface. In this exercise, we calculate it using the cross product of tangent vectors from the parametrized surface equation.

• The tangent vectors are \(\mathbf{r}_u = \langle 1, 0, 0 \rangle\) and \(\mathbf{r}_v = \langle 0, 1, -2v \rangle\), and their cross product is \(\mathbf{n} = \langle 0, 2v, -1 \rangle\).
• This normal vector tells us the direction in which the surfaces point, which impacts how much of the vector field is "flowing through" the surface.

Having the correct orientation is necessary for accurate flux calculations. A normal vector pointing outward from the surface ensures that the region being considered is correctly analyzed in relation to the surrounding space. Awareness of surface direction is vital in flux integrals, as incorrect orientations can lead to erroneous results.