Problem 17 - Calculus Early Transcendentals

Question

Use the Divergence Theorem to find the flux of F across the surface ? with outward orientation. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} ; \sigma \text { is the surface of the coni- }} \\ {\text { cal solid bounded by } z=\sqrt{x^{2}+y^{2}} \text { and } z=1}\end{array} $$

Step-by-Step Solution

Answer

The flux across the surface is \( \frac{5\pi}{6} \).

Step 1: Understand the Problem

The problem asks us to calculate the flux of the vector field \( \mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \) across a given surface \( \sigma \). \( \sigma \) is the surface of a conical solid bounded by the equation \( z = \sqrt{x^2 + y^2} \) and \( z = 1 \). The Divergence Theorem will be used for this calculation.

Step 2: Write the Divergence Theorem Formula

The Divergence Theorem states that the flux of a vector field \( \mathbf{F} \) across a closed surface \( S \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( S \). Mathematically, \( \int_{S} \mathbf{F} \cdot d\mathbf{A} = \iiint_{V} (abla \cdot \mathbf{F}) \, dV \).

Step 3: Calculate the Divergence of \( \mathbf{F} \)

The divergence of \( \mathbf{F} \) is calculated using \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x} (x^2) + \frac{\partial}{\partial y} (y^2) + \frac{\partial}{\partial z} (z^2) \). Each term simplifies to \( 2x, 2y, \) and \( 2z \) respectively, so the divergence is \( abla \cdot \mathbf{F} = 2x + 2y + 2z \).

Step 4: Determine the Limits of Integration

The volume \( V \) enclosed by \( \sigma \) is a conical solid. For integration, express the limits as: \( 0 \leq z \leq 1 \), \( 0 \leq r \leq z \), and \( 0 \leq \theta \leq 2\pi \). Here, \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \).

Step 5: Express Divergence in Cylindrical Coordinates

In cylindrical coordinates, the divergence \( 2x + 2y + 2z \) becomes \( 2r\cos(\theta) + 2r\sin(\theta) + 2z = 2r + 2z \), because \( \cos^2(\theta) + \sin^2(\theta) = 1 \).

Step 6: Set Up and Evaluate the Integral

The flux is given by \( \iiint_{V} (2r + 2z) \, dV \). Substitute the limits and the element \( dV = r \, dr \, d\theta \, dz \). Evaluate: \[ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{z} (2r^2 + 2rz) \, dr \, dz \, d\theta \]. First integrate with respect to \( r \): \[ \int_{0}^{z} (2r^2 + 2rz) \, dr = \left[ \frac{2}{3}r^3 + r^2z \right]_{0}^{z} = \frac{2}{3}z^3 + z^3 = \frac{5}{3}z^3 \]. Then integrate with respect to \( z \): \[ \int_{0}^{1} \frac{5}{3}z^3 \, dz = \left[ \frac{5}{3} \cdot \frac{1}{4}z^4 \right]_{0}^{1} = \frac{5}{12} \]. Finally, integrate with respect to \( \theta \): \[ \int_{0}^{2\pi} \frac{5}{12} \, d\theta = \frac{5}{12} \cdot 2\pi = \frac{5\pi}{6} \].

Step 7: State the Result

The total flux of the vector field \( \mathbf{F} \) across the surface \( \sigma \) is \( \frac{5\pi}{6} \).

Key Concepts

flux calculation

Calculating flux is a central concept in vector calculus, particularly when dealing with vector fields and surfaces in three-dimensional space. Flux refers to the quantity of a vector field that passes through a given surface. To calculate it, we need to evaluate a surface integral of the vector field. By using the Divergence Theorem, we simplify this calculation by transforming it into a volume integral.

The Divergence Theorem connects the flux through a closed surface to the divergence of the vector field over the contained volume. Mathematically, it states:

• \( \int_{S} \mathbf{F} \cdot d\mathbf{A} = \iiint_{V} (abla \cdot \mathbf{F}) \, dV \)

Here, \( \mathbf{F} \) is the vector field, \( S \) is the closed surface, and \( V \) is the enclosed volume. This theorem allows us to transform a two-dimensional problem (surface integral) into a simpler three-dimensional problem (volume integral) by calculating the divergence \( abla \cdot \mathbf{F} \) of the vector field. Hence, rather than directly performing the often complex surface integral, we instead compute the divergence and use triple integration over the volume.

vector field

A vector field is a function that assigns a vector to every point in a subset of space. In simpler terms, you can think of a vector field as a way of describing a space filled with vectors, each having both direction and magnitude.

In the given exercise, the vector field is defined as \( \mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} \). This means that at every point \((x, y, z)\) in space, the vector is \((x^2, y^2, z^2)\). Each component of the vector (\(x^2\), \(y^2\), \(z^2\)) indicates how the vector field behaves along the respective axes in three-dimensional space.

To analyze and utilize the vector field effectively, we often calculate its divergence, \( abla \cdot \mathbf{F} \), which is a scalar field. The divergence gives us an understanding of how the vector field converges or diverges at different points in space. It's calculated as:

• \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x} (x^2) + \frac{\partial}{\partial y} (y^2) + \frac{\partial}{\partial z} (z^2) = 2x + 2y + 2z \)

This expression tells us how much the vector field is expanding or compressing at each point. A positive divergence indicates a source, while a negative divergence can indicate a sink in the vector field.

cylindrical coordinates

Cylindrical coordinates are an extension of the two-dimensional polar coordinates into three dimensions by adding a height axis. They are particularly useful in dealing with problems possessing rotational symmetry, like conical or cylindrical volumes.

In the context of this exercise, converting from Cartesian coordinates (\(x, y, z\)) to cylindrical coordinates (\(r, \theta, z\)) greatly simplifies the integration process. Here is how the transformation works:

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

With these conversions, any calculation over cylindrical surfaces or volumes becomes more manageable, especially when setting up integrals.

The volume integral in cylindrical coordinates involves the volume element \( dV = r \, dr \, d\theta \, dz \), which includes the extra \(r\) term to account for the radial distance in polar coordinates. This alignment of integration variables facilitates efficient computation of the integral boundaries and expressions, enabling the accurate calculation of quantities like the flux using the Divergence Theorem. The particular limits used here are \(0 \leq z \leq 1\), \(0 \leq r \leq z\), and \(0 \leq \theta \leq 2\pi\). This ensures the integration encapsulates the conical solid accurately.