Problem 15 - 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 \(D .\) Wedge \(\mathbf{F}=2 x z \mathbf{i}-x y \mathbf{j}-z^{2} \mathbf{k}\) \(D :\) The wedge cut from the first octant by the plane \(y+z=4\) and the elliptical cylinder \(4 x^{2}+y^{2}=16\)

Step-by-Step Solution

Answer

The outward flux is zero, since the integrals cancel out.

Step 1: Understand the Problem

We are asked to find the outward flux of the vector field \( \mathbf{F}=2 x z \mathbf{i} - x y \mathbf{j} - z^{2} \mathbf{k} \) across the boundary of a region \( D \) using the Divergence Theorem. The region \( D \) is defined by the first octant, bounded by the plane \( y + z = 4 \) and the elliptical cylinder \( 4x^2 + y^2 = 16 \).

Step 2: Divergence of the Vector Field

Calculate the divergence of \( \mathbf{F} \, : abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(2xz) + \frac{\partial}{\partial y}(-xy) + \frac{\partial}{\partial z}(-z^2) = 2z - x - 2z = -x. \)

Step 3: Set Up the Integral for the Divergence Theorem

The Divergence Theorem states that the outward flux through the boundary \( \partial D \) of a region \( D \) is equal to the volume integral of the divergence over \( D \). Thus, we need to evaluate:\[\iiint_{D} (-x) \, dV.\]

Step 4: Determine the Limits of Integration

The region \( D \) is bounded by \( 0 \leq x \leq 2 \), \( 0 \leq y \leq \sqrt{16 - 4x^2} \), and \( 0 \leq z \leq 4 - y \). These limits come from solving the elliptical cylinder equation for \( y \) and from the constraint of the plane for \( z \).

Step 5: Evaluate the Triple Integral

We evaluate the integral:\[\int_{0}^{2} \int_{0}^{\sqrt{16 - 4x^2}} \int_{0}^{4-y} (-x) \, dz \, dy \, dx.\]First, integrate with respect to \( z \):\[ \int_{0}^{4-y} (-x) \, dz = -x(4-y).\]Next, integrate with respect to \( y \):\[ \int_{0}^{\sqrt{16 - 4x^2}} -x(4-y) \, dy.\]This yields:\[ -4x y + xy^2 \bigg|_{0}^{\sqrt{16 - 4x^2}} = -4x(\sqrt{16 - 4x^2}) + x(16 - 4x^2).\]Then, integrate with respect to \( x \):\[\int_{0}^{2} \left(-4x\sqrt{16 - 4x^2} + x(16 - 4x^2)\right) \, dx.\]

Step 6: Simplify and Compute the Final Integral

Calculate the simplified form. This results in:\[ -4 \left [ \frac{(16 - 4x^2)^{3/2}}{3} \right ] + \int_{0}^{2} 16x - 4x^3 \, dx\]Breaking it into parts and integrating leads to:\[ -4 \cdot \frac{1}{3} \cdot 0 + \left[ 8x^2 - x^4 \right]_0^2\]Evaluating at the boundaries the left with:\[= 8(4) - (16) - (0) \]

Key Concepts

Vector Field

A vector field is a function that assigns a vector to every point in a subset of space. It's often denoted by \( \mathbf{F} \), especially in three-dimensional space. In this exercise, the vector field is given by \( \mathbf{F}=2 x z \mathbf{i} - x y \mathbf{j} - z^{2} \mathbf{k} \).
Understanding vector fields is crucial as they model physical quantities that have both magnitude and direction, such as forces or velocities in physics.
Vectors consist of three components along the x, y, and z axes, and each component is a function of these coordinates.

• \( 2 x z \mathbf{i} \) implies a vector component along the x-axis. Its magnitude depends on both \( x \) and \( z \).
• \( -x y \mathbf{j} \) implies a vector component along the y-axis, dependent on \( x \) and \( y \).
• \( -z^{2} \mathbf{k} \) implies a vector component along the z-axis, dependent solely on \( z \).

Understanding each part of the vector field helps in calculating the divergence, which tells us how much the vector field spreads out or converges at a point in space.

Outward Flux

Outward flux refers to how much of the vector field passes out of a closed surface, which in this case is the boundary of region \( D \).
It gives an idea of how the field behaves at the boundary, often related to flow or force in physical applications.
The Divergence Theorem connects a triple integral over a volume with a surface integral over the boundary of that volume, making calculations easier by simplifying complex surface integrals to volume integrals inside the region.
The outward flux is computed by taking the volume integral of the divergence of the vector field over the specified region \( D \), and it represents the net rate at which the vector field flows across the region's boundary from the inside.

Triple Integral

Triple integrals are used to integrate over three-dimensional regions. They extend the concept of double integrals to 3D.
In this exercise, the triple integral is set up to find the divergence of the vector field over the region \( D \):\[\iiint_{D} (-x) \, dV.\]
This integration involves multiple steps, each over one of the three dimensions.
We first find limits for \( x \), \( y \), and \( z \), based on the description of the region \( D \).

• \( 0 \leq x \leq 2 \) is the range for \( x \).
• For \( y \), \( 0 \leq y \leq \sqrt{16 - 4x^2} \), solved from the elliptical cylinder equation.
• The limit for \( z \) is \( 0 \leq z \leq 4 - y \), following the plane's equation.

These integrals allow us to calculate quantities like volume or outward flux by breaking down the complex three-dimensional structure into manageable parts.

Elliptical Cylinder

An elliptical cylinder is a three-dimensional shape that extends infinitely along one axis, in this case the z-axis, with an elliptical cross-section on perpendicular planes.
Its equation in the problem is \( 4x^2 + y^2 = 16 \), representing an ellipse in the x-y plane.
The elliptical boundary helps define the contour of the region \( D \) from which the wedge is cut.
Like a circular cylinder, an elliptical cylinder’s volume and shape are crucial in determining the limits of integration for the region \( D \). The elliptical cross-section results in variable limits for the \( y \) dimension dependent on \( x \).

• The semi-major axis along x is \( \sqrt{4} = 2 \).
• The semi-minor axis along y is \( \sqrt{16} = 4 \).

Understanding the geometry is vital to setting up the integral and successfully applying the Divergence Theorem.