Evaluate Triple Integral Divergence Of Vector Field F

#title: Evaluating the Triple Integral of Divergence of Vector Field F

In this comprehensive article, we will delve into the process of evaluating the triple integral of the divergence of a vector field. Specifically, we will focus on the vector field ${ \vec{F} = 4x\hat{i} - 2y^2\hat{j} + z^2\hat{k} }$ and the region ${ V }$ bounded by the cylinder ${ x^2 + y^2 = 9 }$, and the planes ${ z = 0 }$ and ${ z = 3 }$. This problem elegantly combines vector calculus with multivariable integration, offering a rich learning experience.

Problem Statement

Evaluate the value of the triple integral ${ \iiint_V \text{div} \vec{F} \, dV }$, where the vector field ${ \vec{F} }$ is given by:

${ \vec{F} = 4x\hat{i} - 2y^2\hat{j} + z^2\hat{k} }$

and the volume ${ V }$ is the region bounded by:

• The cylinder: ${ x^2 + y^2 = 9 }$
• The plane: ${ z = 0 }$
• The plane: ${ z = 3 }$

Understanding the Key Concepts

Before we dive into the solution, let's solidify our understanding of the core concepts involved. This will not only aid in solving this particular problem but also enhance our grasp of vector calculus in general.

1. Divergence of a Vector Field

The divergence of a vector field, denoted as ${ \text{div} \vec{F} }$, is a scalar function that measures the “outward flux” of the vector field at a given point. In simpler terms, it quantifies how much the vector field is expanding or diverging away from that point. For a vector field ${ \vec{F} = P\hat{i} + Q\hat{j} + R\hat{k} }$, the divergence is calculated as:

${ \text{div} \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} }$

The divergence is a fundamental concept in fluid dynamics, electromagnetism, and other areas of physics and engineering. A positive divergence indicates a source (outflow), a negative divergence indicates a sink (inflow), and a zero divergence indicates that the field is incompressible (neither expanding nor contracting).

2. Triple Integrals

A triple integral is an extension of the definite integral to three dimensions. It is used to calculate the integral of a function over a three-dimensional region. In our case, we will be integrating the divergence of the vector field over the volume ${ V }$. Triple integrals can be evaluated in Cartesian, cylindrical, or spherical coordinates, depending on the geometry of the region of integration. The general form of a triple integral is:

${ \iiint_V f(x, y, z) \, dV }$

where ${ f(x, y, z) }$ is the function to be integrated, and ${ dV }$ represents the volume element. In Cartesian coordinates, ${ dV = dx \, dy \, dz }$. In cylindrical coordinates, ${ dV = r \, dr \, d\theta \, dz }$, and in spherical coordinates, ${ dV = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi }$.

3. Cylindrical Coordinates

Given the cylindrical symmetry of the region ${ V }$ (bounded by the cylinder ${ x^2 + y^2 = 9 }$), it is advantageous to use cylindrical coordinates to simplify the triple integral. Cylindrical coordinates ${ (r, \theta, z) }$ are related to Cartesian coordinates ${ (x, y, z) }$ by:

${ x = r \cos(\theta)\\ y = r \sin(\theta)\\ z = z }$

The Jacobian determinant for the transformation from Cartesian to cylindrical coordinates is ${ r }$, which appears in the volume element ${ dV = r \, dr \, d\theta \, dz }$. This transformation often simplifies integrals over cylindrical regions, as it aligns the coordinate system with the geometry of the region.

4. The Divergence Theorem (Optional, but insightful)

While not strictly necessary for solving this problem directly, the Divergence Theorem provides a powerful connection between the triple integral of the divergence of a vector field and the surface integral of the vector field over the boundary of the region. The Divergence Theorem states:

${ \iiint_V \text{div} \vec{F} \, dV = \oiint_{\partial V} \vec{F} \cdot d\vec{S} }$

where ${ \partial V }$ is the boundary surface of the volume ${ V }$, and ${ d\vec{S} }$ is the outward-pointing normal vector element on the surface. Understanding the Divergence Theorem can offer alternative perspectives on problems involving divergences and fluxes, and sometimes provides a more efficient method of calculation.

Step-by-Step Solution

Now, let's proceed with the solution step-by-step:

Step 1: Calculate the Divergence of ${ \vec{F} }$

Given the vector field ${ \vec{F} = 4x\hat{i} - 2y^2\hat{j} + z^2\hat{k} }$, we first compute its divergence using the formula:

${ \text{div} \vec{F} = \frac{\partial}{\partial x}(4x) + \frac{\partial}{\partial y}(-2y^2) + \frac{\partial}{\partial z}(z^2) }$

Taking the partial derivatives, we get:

${ \frac{\partial}{\partial x}(4x) = 4 }$

${ \frac{\partial}{\partial y}(-2y^2) = -4y }$

${ \frac{\partial}{\partial z}(z^2) = 2z }$

Therefore, the divergence of ${ \vec{F} }$ is:

${ \text{div} \vec{F} = 4 - 4y + 2z }$

Step 2: Set up the Triple Integral in Cylindrical Coordinates

Since the region ${ V }$ is a cylinder, we will convert the triple integral into cylindrical coordinates. The cylindrical coordinates are defined as:

${ x = r \cos(\theta)\\ y = r \sin(\theta)\\ z = z }$

with the volume element ${ dV = r \, dr \, d\theta \, dz }$. We need to express the divergence in cylindrical coordinates as well. Substituting ${ y = r \sin(\theta) }$ into the divergence expression, we get:

${ \text{div} \vec{F} = 4 - 4(r \sin(\theta)) + 2z = 4 - 4r \sin(\theta) + 2z }$

Now, we determine the limits of integration. The region ${ V }$ is bounded by:

• The cylinder ${ x^2 + y^2 = 9 }$, which in cylindrical coordinates is ${ r^2 = 9 }$ or ${ r = 3 }$. So, ${ 0 \leq r \leq 3 }$.
• The angle ${ \theta }$ covers the entire circle, so ${ 0 \leq \theta \leq 2\pi }$.
• The planes ${ z = 0 }$ and ${ z = 3 }$, so ${ 0 \leq z \leq 3 }$.

Thus, the triple integral in cylindrical coordinates is:

${ \iiint_V \text{div} \vec{F} \, dV = \int_{0}^{3} \int_{0}^{2\pi} \int_{0}^{3} (4 - 4r \sin(\theta) + 2z) r \, dz \, d\theta \, dr }$

Step 3: Evaluate the Integral

We now evaluate the triple integral step-by-step.

First, integrate with respect to ${ z }$:

${ \int_{0}^{3} (4 - 4r \sin(\theta) + 2z) r \, dz = r \int_{0}^{3} (4 - 4r \sin(\theta) + 2z) \, dz }$

${ = r \left[ 4z - 4r \sin(\theta) z + z^2 \right]_{0}^{3} }$

${ = r \left[ 4(3) - 4r \sin(\theta) (3) + (3)^2 \right] = r (12 - 12r \sin(\theta) + 9) = r (21 - 12r \sin(\theta)) }$

So, the integral becomes:

${ \int_{0}^{2\pi} \int_{0}^{3} r (21 - 12r \sin(\theta)) \, dr \, d\theta }$

Next, integrate with respect to ${ r }$:

${ \int_{0}^{3} r (21 - 12r \sin(\theta)) \, dr = \int_{0}^{3} (21r - 12r^2 \sin(\theta)) \, dr }$

${ = \left[ \frac{21}{2} r^2 - 4r^3 \sin(\theta) \right]_{0}^{3} = \frac{21}{2} (3)^2 - 4(3)^3 \sin(\theta) = \frac{21}{2} (9) - 4(27) \sin(\theta) = \frac{189}{2} - 108 \sin(\theta) }$

Now, the integral is:

${ \int_{0}^{2\pi} \left( \frac{189}{2} - 108 \sin(\theta) \right) d\theta }$

Finally, integrate with respect to ${ \theta }$:

${ \int_{0}^{2\pi} \left( \frac{189}{2} - 108 \sin(\theta) \right) d\theta = \left[ \frac{189}{2} \theta + 108 \cos(\theta) \right]_{0}^{2\pi} }$

${ = \frac{189}{2} (2\pi) + 108 \cos(2\pi) - \left( \frac{189}{2} (0) + 108 \cos(0) \right) }$

${ = 189\pi + 108 - 108 = 189\pi }$

Therefore, the value of the triple integral is:

${ \iiint_V \text{div} \vec{F} \, dV = 189\pi }$

Conclusion

We have successfully evaluated the triple integral of the divergence of the vector field ${ \vec{F} = 4x\hat{i} - 2y^2\hat{j} + z^2\hat{k} }$ over the region ${ V }$ bounded by the cylinder ${ x^2 + y^2 = 9 }$ and the planes ${ z = 0 }$ and ${ z = 3 }$. By first computing the divergence, setting up the triple integral in cylindrical coordinates, and then evaluating the integral iteratively, we arrived at the result ${ 189\pi }$. This problem demonstrates the power of vector calculus and multivariable integration in solving real-world problems in physics and engineering. The use of cylindrical coordinates significantly simplified the integral due to the symmetry of the region, highlighting the importance of choosing the right coordinate system for a given problem. Understanding the Divergence Theorem provides an additional layer of insight into the relationship between volume integrals of divergence and surface integrals of flux, reinforcing the interconnectedness of these concepts in vector calculus.