Problem 398 - Calculus Volume 3

Question

For the following exercises, use a CAS along with the divergence theorem to compute the net outward flux for the fields across the given surfaces \(S .\) \(\mathbf{I T} \mathbf{I} \mathbf{F}=\langle x, y, z\rangle ; S\) is the surface of paraboloid \(z=4-x^{2}-y^{2}, \quad\) for \(\quad z \geq 0,\) plus its base in the \(x y\) -plane.

Step-by-Step Solution

Answer

The net outward flux is \( 16\pi \).

Step 1: Understand the Problem

We need to calculate the net outward flux of the vector field \( \mathbf{F} = \langle x, y, z \rangle \) across the surface of a paraboloid \( z = 4 - x^2 - y^2 \) with its base on the \( xy \)-plane. We will use the divergence theorem for this calculation.

Step 2: Identify the Volume V

The region \( V \) is the volume enclosed by the paraboloid \( z = 4 - x^2 - y^2 \) and the \( xy \)-plane that forms a circular paraboloid cap with radius 2, as \( z = 0 \) intersects the \( xy \)-plane at a circle of radius 2.

Step 3: Divergence of the Field

Calculate the divergence of the vector field \( \mathbf{F} = \langle x, y, z \rangle \). The divergence \( abla \cdot \mathbf{F} \) is obtained by calculating the partial derivatives: \( \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = 3 \).

Step 4: Setup the Volume Integral

According to the divergence theorem, the net outward flux \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \) is equal to \( \iiint_V abla \cdot \mathbf{F} \, dV \). Thus, we need to integrate 3 over the volume \( V \), where \( V \) is bounded by the paraboloid and the \( xy \)-plane.

Step 5: Convert to Cylindrical Coordinates

Convert the problem to cylindrical coordinates, where \( x = r\cos\theta \), \( y = r\sin\theta \), and \( z = z \). The bounds become \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq z \leq 4 - r^2 \).

Step 6: Evaluate the Integral

Integrate \( 3 \) over the volume in cylindrical coordinates: \[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4-r^2} 3 \, r \, dz \, dr \, d\theta. \] This evaluates to \( \int_{0}^{2\pi} \int_{0}^{2} [3rz]_{0}^{4-r^2} \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{2} 3r(4-r^2) \, dr \, d\theta \). Work through the integrals step-by-step to find the flux.

Step 7: Calculate the Final Result

First, integrate with respect to \( z \): \[ 3r \times (4-r^2) = 12r - 3r^3. \] Then integrate \( \int_{0}^{2} (12r - 3r^3) \, dr \), followed by \( \int_{0}^{2\pi} \, d\theta \). The final evaluation of the nested integral gives \( 16\pi \).

Key Concepts

Net Outward Flux

Net outward flux is a measure of how much a vector field flows out of a given closed surface. Imagine a fluid moving through a surface; the net outward flux would represent the total quantity of fluid that exits through this surface. This concept is crucial in understanding fluid mechanics and electrical fields.
To calculate the net outward flux mathematically, we use surface integrals. For vector fields, this involves integrating the vector field over a surface and is often represented as \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{F} \) is the vector field, and \( \mathbf{n} \) is the outward normal vector to the surface \( S \).
The computation can be simplified using the Divergence Theorem. This theorem connects a surface integral with a volume integral, allowing us to use easier-to-handle integrals to find the flux.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional extension of polar coordinates, useful for problems with symmetry about an axis, like cylindrical or paraboloid shapes. They consist of three components: \( r \), the radial distance from the axis; \( \theta \), the angular position around the axis; and \( z \), the height from a reference plane. These components offer a natural fit for objects like the paraboloid in the given exercise.
To convert from Cartesian coordinates to cylindrical ones, use the transformations:

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

When evaluating integrals, the differential volume element becomes \( r \, dz \, dr \, d\theta \), incorporating the radius \( r \) since it varies with \( \theta \) and \( z \). This conversion simplifies the bounds of the integral and makes computations more manageable.

Vector Field Integration

Vector field integration involves calculating the flow of a vector field through a surface. In the context of the exercise, this involves integrating the vector field \( \mathbf{F} = \langle x, y, z \rangle \) across the surface of a paraboloid.
This process generally requires calculating both the surface and volume integrals, but with the Divergence Theorem, the surface integral \( \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \) can be replaced with a volume integral \( \iiint_V abla \cdot \mathbf{F} \, dV \).
The divergence function \( abla \cdot \mathbf{F} \) is the measure of the tendency of the vector field to originate from or converge into a point within the volume. For this exercise, it simplifies to \( \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 3 \).
Using this divergence simplifies the vector field integration into an easier volume integral.

Volume Integral Calculation

Calculating a volume integral in this context involves evaluating an integral over a specified three-dimensional region. Here, the region is the volume bounded by the paraboloid \( z = 4 - x^2 - y^2 \) and the \( xy \)-plane.
Translating this into cylindrical coordinates gives us the bounds for \( r, \theta, \) and \( z \):

• \( 0 \leq r \leq 2 \)
• \( 0 \leq \theta \leq 2\pi \)
• \( 0 \leq z \leq 4 - r^2 \)

The integral then becomes \( \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{4-r^2} 3r \, dz \, dr \, d\theta \). This evaluates the divergence, creating a simpler route to calculating the flux compared to traditional methods.
Breaking down the integral calculation, we evaluate from the innermost to the outermost part:

• Integrate with respect to \( z \), resulting in \( [3rz]_{0}^{4-r^2} = 3r(4-r^2) \)
• Integrate \( 3r(4-r^2) \) with respect to \( r \), leading to \( 12r - 3r^3 \)
• Finally, integrate with respect to \( \theta \) over its full range from 0 to \( 2\pi \)

The final outcome of these calculations yields the net outward flux, illustrating the effectiveness of the divergence theorem.