Problem 11 - Thomas Calculus

Question

In Exercises 9-20, use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) Cylinder and paraboloid \(\mathbf{F}=y \mathbf{i}+x y \mathbf{j}-z \mathbf{k}\) \(D :\) The region inside the solid cylinder \(x^{2}+y^{2} \leq 4\) between the plane \(z=0\) and the paraboloid \(z=x^{2}+y^{2}\)

Step-by-Step Solution

Answer

The outward flux is \(-8\pi\).

Step 1: Understand the Problem and Identify the Region

The problem involves finding the outward flux of the vector field \(\mathbf{F} = y \mathbf{i} + x y \mathbf{j} - z \mathbf{k}\) across the boundary of a region \(D\). The region \(D\) is defined as the volume inside a cylinder \(x^2 + y^2 \leq 4\) above the plane \(z=0\) and below the paraboloid \(z = x^2 + y^2\).

Step 2: Apply the Divergence Theorem

The Divergence Theorem states that the flux of a vector field \(\mathbf{F}\) across a closed surface is equal to the triple integral of the divergence of \(\mathbf{F}\) over the volume \(D\). Mathematically, this is expressed as: \[ \iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_D abla \cdot \mathbf{F} \, dV \] where \(\partial D\) is the boundary of \(D\) and \(abla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\).

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

The divergence of \(\mathbf{F} = y \mathbf{i} + x y \mathbf{j} - z \mathbf{k}\) is calculated as: \[ abla \cdot \mathbf{F} = \frac{\partial y}{\partial x} + \frac{\partial (xy)}{\partial y} + \frac{\partial (-z)}{\partial z} \]Evaluating each term gives:- \(\frac{\partial y}{\partial x} = 0\)- \(\frac{\partial (xy)}{\partial y} = x\)- \(\frac{\partial (-z)}{\partial z} = -1\)Thus, \(abla \cdot \mathbf{F} = 0 + x - 1 = x - 1\).

Step 4: Set Up the Triple Integral

The integral becomes: \[ \iiint_D (x - 1) \, dV \]We express this integral in cylindrical coordinates since the region \(D\) is bound by a cylinder, translated to: \((r, \theta, z)\) with bounds for \(r\): 0 to 2, for \(\theta\): 0 to \(2\pi\), and for \(z\): 0 to \(r^2\). Thus, the integral is:\[ \int_0^{2\pi} \int_0^2 \int_0^{r^2} (r \cos \theta - 1)r \, dz \, dr \, d\theta \]

Step 5: Solve the Triple Integral

Evaluating the integral step by step:1. Integrate with respect to \(z\): \[ \int_0^{r^2} (r^2\cos \theta - r) \, dz = r^2\cos \theta \cdot z - r \cdot z \Big|_0^{r^2} = r^4\cos \theta - r^3 \]2. Integrate with respect to \(r\): \[ \int_0^2 (r^4\cos \theta - r^3) \, dr = \frac{r^5}{5}\cos \theta - \frac{r^4}{4} \Big|_0^2 = \left(\frac{32}{5}\cos \theta - 4\right) \]3. Integrate with respect to \(\theta\): \[ \int_0^{2\pi} \left(\frac{32}{5}\cos \theta - 4\right) \, d\theta = \left[\frac{32}{5}\sin \theta\right]_0^{2\pi} - 4\cdot2\pi = 0 - 8\pi \]Thus, the flux is \(-8\pi\).

Step 6: Interpret the Result

The result of \(-8\pi\) indicates that the net outward flux of the vector field \(\mathbf{F}\) across the boundary of region \(D\) is \(-8\pi\). This negative value implies more flow is inward through the boundary than outward.

Key Concepts

Vector Field

A vector field is a mathematical construct that assigns a vector to every point in a space. In this exercise, we are dealing with a specific vector field \( \mathbf{F} = y \mathbf{i} + x y \mathbf{j} - z \mathbf{k} \), which describes a field in three-dimensional space. Here, the components of the vector field are expressed in terms of \( x \), \( y \), and \( z \), with \( y \mathbf{i} \) being the component in the x-direction. This means that at any point \( (x, y, z) \) in space, the vector field will have a direction and magnitude based on these equations.Understanding how a vector field behaves is crucial when working with concepts like flux and the Divergence Theorem. In simple terms, vector fields can represent a variety of phenomena, like the flow of fluid or the force acting on a charged particle. To grasp vector fields fully:

• The component \( y \mathbf{i} \) suggests movement parallel to the x-axis, modifying in the y-direction.
• The component \( x y \mathbf{j} \) indicates an effect along the y-axis, influenced by both x and y values.
• The component \( -z \mathbf{k} \) implies a downward direction in terms of the z-axis.

Visualizing how vectors change across space is vital to understanding how flux behaves in a vector field.

Outward Flux

Outward flux is a measure of how much of the vector field exits, or "flows" out of, a given surface. When considering the Divergence Theorem, which connects the external flow through a surface to the behavior within a volume, understanding outward flux is paramount. It is computed as the surface integral of the vector field dotting with the unit normal vector \( \mathbf{n} \).In this particular problem, we calculate the outward flux of \( \mathbf{F} \) across the boundary of region \( D \). A positive flux typically indicates that the vector field vectors are mostly pointing outwards, while a negative flux, such as the \( -8\pi \) obtained in the solution, indicates more vectors are pointing inward.To comprehend outward flux:

• The flux quantifies the amount of the field leaving through the surface.
• Negative values suggest net inward flow, implying more input into a surface than output.
• Flux is essential for evaluating conservation laws, like how fluid naturally exits or enters a boundary.

Analyzing flux helps you determine the tendency of a field towards inward or outward flow.

Cylindrical Coordinates

Cylindrical coordinates are a powerful coordinate system used to simplify the calculation of integrals when symmetries are circular or cylindrical in nature. In cylindrical coordinates, positions in space are described using radial distance \( r \), angle \( \theta \), and height \( z \).In this exercise, we apply cylindrical coordinates to build the integral for calculating flux:

• Transformation from Cartesian coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \), \( z = z \).
• Suitable for problems with circular symmetry, such as cylinders.
• Simplifies triple integrals by aligning the coordinate system with the geometry of the problem.

Using cylindrical coordinates allows for a reduced complexity when evaluating integrals over cylindrical volumes, making it a very efficient approach for problems involving cylinders.

Solid Cylinder

A solid cylinder is a three-dimensional geometric shape characterized by a circular base and a height. It is an essential structure in physics and engineering, playing a role in understanding volumes and surface assessments. The exercise problem deals with a solid cylinder characterized by \( x^2 + y^2 \leq 4 \), meaning each circular cross-section has a radius of 2.The volume you're considering is bounded above by a paraboloid \( z = x^2 + y^2 \) and below by the plane \( z = 0 \). To conceptualize a solid cylinder:

• The circular base is perpendicular to the rotational axis, with height extending upwards.
• The side of the cylinder, or curved surface, wraps around the structure from base to base.
• The intersection of such a form with other surfaces, like paraboloids, defines unique volumes to analyze flow and flux in vector fields.

Understanding the parameters and boundaries of a solid cylinder are key when applying the Divergence Theorem to analyze outward flux through its surface.