Problem 14 - 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 \(D .\) Cylindrical can \(\mathbf{F}=\left(6 x^{2}+2 x y\right) \mathbf{i}+\left(2 y+x^{2} z\right) \mathbf{j}+4 x^{2} y^{3} \mathbf{k}\) \(D :\) The region cut from the first octant by the cylinder \(x^{2}+y^{2}=4\) and the plane \(z=3\)

Step-by-Step Solution

Answer

The outward flux across the boundary of \( D \) is 44.

Step 1: Understand the Region

We need to find the outward flux of \( \mathbf{F} \) across the boundary of region \( D \), which is a cylindrical can found in the first octant, bounded by the cylinder \( x^2 + y^2 = 4 \) and the plane \( z = 3 \). As the can is in the first octant, \( x, y, z \geq 0 \).

Step 2: Apply Divergence Theorem

The Divergence Theorem states that for a vector field \( \mathbf{F} \) and a region \( D \), the flux of \( \mathbf{F} \) through the boundary \( \partial D \) is equal to the triple integral over \( D \) of the divergence of \( \mathbf{F} \). Mathematically, \( \int_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \int_{D} abla \cdot \mathbf{F} \, dV \).

Step 3: Calculate Divergence

Compute \( abla \cdot \mathbf{F} \) where \( \mathbf{F} = (6x^2 + 2xy)\mathbf{i} + (2y + x^2z)\mathbf{j} + 4x^2y^3\mathbf{k} \). The divergence is given by: \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(6x^2 + 2xy) + \frac{\partial}{\partial y}(2y + x^2z) + \frac{\partial}{\partial z}(4x^2y^3) \).

Step 4: Evaluate Partial Derivatives

Calculate each partial derivative: - \( \frac{\partial}{\partial x}(6x^2 + 2xy) = 12x + 2y \)- \( \frac{\partial}{\partial y}(2y + x^2z) = 2 \)- \( \frac{\partial}{\partial z}(4x^2y^3) = 0 \)

Step 5: Integrate Over the Region

Combine the partial derivatives to find \( abla \cdot \mathbf{F} = 12x + 2y + 2 \). Now, we integrate this over the region \( D \). The region \( D \) can be parameterized in cylindrical coordinates as \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq \frac{\pi}{2} \), and \( 0 \leq z \leq 3 \).

Step 6: Convert to Cylindrical Coordinates and Integrate

Convert \( abla \cdot \mathbf{F} \) to cylindrical coordinates. Since \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), we have: - \( 12x = 12r\cos(\theta) \)- \( 2y = 2r\sin(\theta) \)The integral becomes \( \int_{0}^{3} \int_{0}^{2} \int_{0}^{\pi/2} \left(12r \cos(\theta) + 2r \sin(\theta) + 2\right) r \, d\theta \, dr \, dz \), where \( r \, d\theta \, dr \, dz \) is the Jacobian of the transformation.

Step 7: Solve the Integral

Compute the integral step-by-step:1. Integrate with respect to \( \theta \):\[ \int_{0}^{\pi/2} \left(12r^2 \cos(\theta) + 2r^2 \sin(\theta) + 2r \right) \ d\theta \]2. Integrate with respect to \( r \) from 0 to 2.3. Integrate with respect to \( z \) from 0 to 3.

Step 8: Final Calculation

Carry out the remaining calculations from Step 7 to arrive at the final value. The triple integral evaluates to the total flux across the boundary of \( D \).

Key Concepts

Vector Field

A vector field is a crucial mathematical concept that allows us to assign a vector to every point in space. Imagine arrows at different locations that have both direction and magnitude. In the exercise, the vector field is given by \(\mathbf{F} = (6x^2 + 2xy)\mathbf{i} + (2y + x^2z)\mathbf{j} + 4x^2y^3\mathbf{k}\). This vector field represents different flows across the space defined by variables \(x\), \(y\), and \(z\).

Here’s how each component works:

• \(6x^2 + 2xy\) controls the direction and magnitude in the \(x\)-axis or \(\mathbf{i}\)-direction.
• \(2y + x^2z\) affects the \(y\)-axis or \(\mathbf{j}\)-direction.
• \(4x^2y^3\) governs the \(z\)-axis or \(\mathbf{k}\)-direction.

Understanding how each component works allows you to visualize how different regions have varying intensities and directions of the vector field within the cylindrical can region.

Flux Calculation

Calculating the flux involves understanding how the vector field passes through a surface. Flux is essentially the flow rate of the vector field across a boundary, like a river flowing through a net. The Divergence Theorem simplifies flux calculations by equating the surface integral of a vector field to a volume integral of its divergence.

For flux calculation, the problem instructs to use the Divergence Theorem, which states: \[\int_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \int_{D} abla \cdot \mathbf{F} \, dV\]
In this formula, \(abla \cdot \mathbf{F}\) is the divergence, which measures how much the vector field spreads out from a point. By converting the problem using the Divergence Theorem, we transition from complex surface calculations to more straightforward volume calculations.

Triple Integral

The triple integral is a powerful tool in mathematics used for calculating volume-associated properties over three-dimensional regions. Here, it helps evaluate the total flux through the cylindrical can.

To approach this, we first calculate the divergence of \(\mathbf{F}\) by finding:

• \(\frac{\partial}{\partial x}(6x^2 + 2xy) = 12x + 2y\)
• \(\frac{\partial}{\partial y}(2y + x^2z) = 2\)
• \(\frac{\partial}{\partial z}(4x^2y^3) = 0\)

The divergence \(abla \cdot \mathbf{F}\) becomes \(12x + 2y + 2\).

Next, we integrate this divergence over the volume defined by the given parameters: height from 0 to 3, radial distance from 0 to 2, and angle from 0 to \(\pi/2\). This converts our problem into a triple integral where \(r \, d\theta \, dr \, dz\) is the volume element.

Cylindrical Coordinates

Cylindrical coordinates simplify problems involving symmetrical shapes like cylinders. They transform \(x\), \(y\), and \(z\) into \((r, \theta, z)\), making it easier to solve integrals in regions with rotational symmetry.

The transformations are:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)
• \(z = z\) remains the same as it is aligned with the axis of symmetry.

In our exercise, using cylindrical coordinates helps express the shape of the region \(D\) more naturally.
By expressing \(abla \cdot \mathbf{F}\) in terms of \(r\) and \(\theta\), such as

• \(12x = 12r\cos(\theta)\)
• \(2y = 2r\sin(\theta)\)

we streamline our calculations, ensuring our integral calculations align with the geometry of the cylindrical can.