Problem 28 - 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. Paraboloid \(\mathbf{F}=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k}\) outward (normal away from the \(z\) -axis) through the surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=1\)

Step-by-Step Solution

Answer

The flux across the surface is \(-4\pi\).

Step 1: Parametrize the Surface

The equation of the paraboloid is given by \(z = x^2 + y^2\). To parametrize the surface, notice that it is a surface of revolution around the z-axis. We choose parameters \(u\) and \(v\) as follows: \(x = u\cos(v), \ y = u\sin(v)\) with \(0 \leq u \leq 1\) and \(0 \leq v < 2\pi\). Thus, \(z = u^2\) when restricted by the plane \(z=1\), which implies \(u^2 = 1\), so \(u = 1\). Hence, the parametrization becomes \(\mathbf{r}(v) = \langle \cos(v), \sin(v), 1 \rangle\).

Step 2: Compute the Normal Vector

To find the flux, we need a normal vector to the surface. The vector differential area element is given by \(\frac{\partial \mathbf{r}}{\partial v} \times \frac{\partial \mathbf{r}}{\partial u}\). From our parametrization, we have \(\frac{\partial \mathbf{r}}{\partial v} = \langle -\sin(v), \cos(v), 0 \rangle\) and the radial vector is \(\langle -2x, -2y, 1 \rangle\). Evaluating at \(u=1\), the normal vector is \(\mathbf{n} = \langle 2, 2, -1 \rangle\) due to orientation conditions (outward normal).

Step 3: Calculate the Flux Across the Surface

The flux is calculated by the surface integral: \(\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma\). Here, \(\mathbf{F} = 4x\mathbf{i} + 4y\mathbf{j} + 2\mathbf{k}\). Substitute \(\mathbf{F} = \langle 4\cos(v), 4\sin(v), 2 \rangle\) and \(\mathbf{n} = \langle 2, 2, -1 \rangle\). Thus, \(\mathbf{F} \cdot \mathbf{n} = (4\cos(v) \cdot 2) + (4\sin(v) \cdot 2) + (2 \cdot -1) = 8\cos(v) + 8\sin(v) - 2\).

Step 4: Integrate Over the Domain

Now, integrate \(\mathbf{F} \cdot \mathbf{n}\) over the region. The integral becomes \(\int_{0}^{2\pi} (8\cos(v) + 8\sin(v) - 2) \, dv\). Compute each integral separately: \(\int_{0}^{2\pi} 8\cos(v) \, dv = 0\), \(\int_{0}^{2\pi} 8\sin(v) \, dv = 0\), and \(\int_{0}^{2\pi} -2 \, dv = -4\pi\). The total flux is \(-4\pi\).

Key Concepts

Surface Parametrization

Surface parametrization is a technique used to describe a surface in a space using parameters. This approach is useful for calculating flux, among other things.

Here, a paraboloid is parametrized. A paraboloid is defined by the equation \( z = x^2 + y^2 \). This surface can be visualized as a bowl centered on the z-axis. Parametrization involves defining coordinate variables \( x \) and \( y \) in terms of parameters \( u \) and \( v \).

For this problem, the surface is described using polar coordinates, where:

• \( x = u \cos(v) \)
• \( y = u \sin(v) \)

Given the condition where the plane \( z = 1 \) intersects the paraboloid, \( u^2 = 1 \) implies \( u = 1 \).

Thus, the parametrization of the surface becomes \( \mathbf{r}(v) = \langle \cos(v), \sin(v), 1 \rangle \). This effectively captures all points on the paraboloid's surface intersecting the plane for later calculations.

Normal Vector Calculation

To calculate flux, a critical step is determining the normal vector. The normal vector gives the direction perpendicular to the surface at points specified by our parametrization. This vector is crucial for surface integrals as it influences the calculation of flux across the surface.

From the parametrized vector \( \mathbf{r}(v) = \langle \cos(v), \sin(v), 1 \rangle \), we find derivatives with respect to the parameters.

• \( \frac{\partial \mathbf{r}}{\partial v} = \langle -\sin(v), \cos(v), 0 \rangle \)

For the radial component in our paraboloid, we apply the gradient \( \langle -2x, -2y, 1 \rangle \).

At \( u = 1 \), the outward normal vector can be deduced as \( \mathbf{n} = \langle 2, 2, -1 \rangle \). This vector points away from the z-axis as required in the problem.

Paraboloid

A paraboloid is a three-dimensional surface shaped like an elongated bowl. Its equation is typically of the form \( z = x^2 + y^2 \), which shows how the surface curves upwards from the origin in the 3D space.

For this problem, a section of the paraboloid is considered where it is cut by the plane \( z = 1 \). This intersection creates a circular boundary which simplifies the problem by reducing the area of interest. Visualizing this intersection is key, as it helps us understand why the parametric form \( \mathbf{r}(v) \) is valid for this precise section.

Understanding the paraboloid’s shape and the role of intersecting planes is fundamental to grasping the surface's geometry and how we apply integrals to compute flux.

Surface Integrals

Surface integrals extend the concept of single-variable integrals to functions defined over surfaces. They are applied to quantify the total of a field over a given surface. Here, surface integrals are used to compute the flux of \( \mathbf{F} = 4x\mathbf{i} + 4y\mathbf{j} + 2\mathbf{k} \) across the surface.

This is mathematically represented by the integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \), where \( \mathbf{F} \cdot \mathbf{n} \) is the scalar product of the vector field and the normal vector.

Using our previously calculated normal vector and the given \( \mathbf{F} \) field, \( \mathbf{F} \cdot \mathbf{n} = 8\cos(v) + 8\sin(v) - 2 \) expresses the degree of field flux at each point. Integrating this across the full parameter range \( 0 \leq v < 2\pi \), results in a total flux of \( -4\pi \).

• This process captures cumulative interactions over the entire selected paraboloid section.