Problem 8 - Advanced Engineering Mathematics

Question

Use the divergence theorem to find the outward flux \(\iint_{S}(\mathbf{F} \cdot \mathbf{n}) d S\) of the given vector field \(\mathbf{F}\). \(\mathbf{F}=\left(x^{2}+\sin y\right) \mathbf{i}+z^{2} \mathbf{j}+x y^{3} \mathbf{k} ; D\) the region bounded by $$ y=x^{2}, z=9-y, z=0 $$

Step-by-Step Solution

Answer

The outward flux of the vector field is 0 due to the symmetry over the given region.

Step 1: State the Divergence Theorem

The Divergence Theorem, also known as Gauss's Theorem, states that for a vector field \( \mathbf{F} \) and a closed surface \( S \) that encloses a volume \( D \), the outward flux through \( S \) is equal to the volume integral of the divergence of \( \mathbf{F} \) over \( D \). Mathematically, \( \iint_{S}(\mathbf{F} \cdot \mathbf{n}) \, dS = \iiint_{D} (abla \cdot \mathbf{F}) \, dV \).

Step 2: Calculate the Divergence of the Vector Field

The divergence of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). For the given vector field, \( \mathbf{F}=(x^2 + \sin y)\mathbf{i} + z^2\mathbf{j} + xy^3\mathbf{k} \), we have:- \( P = x^2 + \sin y \), hence \( \frac{\partial P}{\partial x} = 2x \).- \( Q = z^2 \), hence \( \frac{\partial Q}{\partial y} = 0 \).- \( R = xy^3 \), hence \( \frac{\partial R}{\partial z} = 0 \).Therefore, \( abla \cdot \mathbf{F} = 2x + 0 + 0 = 2x \).

Step 3: Set Up the Volume Integral

Now we need to evaluate the triple integral \( \iiint_{D} 2x \, dV \). The region \( D \) is bounded by \( y = x^2 \), \( z = 9 - y \), and \( z = 0 \). This implies:- For \( y \), we have \( y \) ranging from \( x^2 \) to \( 9 \). - For \( z \), it ranges from 0 to \( 9 - y \). - The bounds for \( x \) will be determined by the shape of \( y = x^2 \), ranging approximately from \( -3 \) to \( 3 \) in the \( xy \)-plane intersection.

Step 4: Evaluate the Triple Integral

We compute the integral: \[ \int_{-3}^{3} \int_{x^2}^{9} \int_{0}^{9-y} 2x \, dz \, dy \, dx. \]First, integrate with respect to \( z \):\[ \int_{0}^{9-y} 2x \, dz = 2x(9-y) - 2x(0) = 2x(9-y). \]Next, integrate with respect to \( y \):\[ \int_{x^2}^{9} 2x(9-y) \, dy = 2x \left[ 9y - \frac{y^2}{2} \right]_{x^2}^{9}. \]Evaluate the expression:\[ = 2x \left[ 81 - \frac{81}{2} - (9x^2 - \frac{x^4}{2}) \right]. \]Finally, integrate with respect to \( x \):\[ \int_{-3}^{3} 2x \left[ 81 - \frac{81}{2} - (9x^2 - \frac{x^4}{2}) \right] \, dx. \]

Step 5: Calculation of the Integral

Now integrate the expression:\[ = 2 \int_{-3}^{3} x (81 - \frac{81}{2} - 9x^2 + \frac{x^4}{2}) \, dx. \]Focus on evaluating the odd function contributions from \(-3\) to \(3\), especially since the integral of an odd function over symmetric limits is zero.The terms 81 and \(-9x^2 + \frac{x^4}{2}\) need detailed integration:1. \( \int_{-3}^{3} 81x \, dx = 0 \) by symmetry (odd function).2. \( \int_{-3}^{3} -9x^3 \, dx + \frac{1}{2} \int_{-3}^{3} x^5 \, dx = 0 \) by symmetry.Due to symmetry in integral over \(-3\) to \(3\), each non-zero result cancels out.

Key Concepts

Vector Field

A vector field is a mathematical structure where a vector, consisting of both magnitude and direction, is assigned to every point in space. Imagine a vector field as a collection of arrows spread throughout a region, where each arrow represents the vector at that point.
Vector fields are essential in physics and engineering as they model the behavior of forces in space. For instance, they can represent gravitational, electric, or magnetic fields.
Specifically, in our given problem, we have the vector field \[ \mathbf{F} = (x^2 + \sin y)\mathbf{i} + z^2 \mathbf{j} + xy^3\mathbf{k} \]. This field contains three components: one for each spatial dimension. Each component is a function that may vary with respect to \(x\), \(y\), or \(z\) and affects the vector's behavior at any given point.

• The \(\mathbf{i}\)-component \((x^2 + \sin y)\) implies how the field behaves along the x-axis.
• The \(\mathbf{j}\)-component \((z^2)\) shows the variation along the y-axis.
• The \(\mathbf{k}\)-component \((xy^3)\) determines the field's behavior along the z-axis.

Understanding the nature and transformation of these vectors over space is a crucial step in applying integral theorems.

Volume Integral

The volume integral is a mathematical tool used to integrate a function over a three-dimensional region. Unlike simple integration, which only considers a single variable, volume integration accounts for variations across all three spatial dimensions.
In the context of the Divergence Theorem, the volume integral of the divergence of a vector field allows us to compute the total divergence within a volume. For our problem, we determined the divergence to be \(abla \cdot \mathbf{F} = 2x\).
The region \(D\) bounded by \(y = x^2\), \(z = 9 - y\), and \(z = 0\), is explored by setting limits on \(y\) and \(z\):

• \(y\) ranges from \(x^2\) to \(9\).
• \(z\) ranges from 0 to \(9 - y\).
• \(x\) ranges approximately from \(-3\) to \(3\).

Setting up the triple integral:
\[\iiint_{D} 2x \, dV = \int_{-3}^{3} \int_{x^2}^{9} \int_{0}^{9-y} 2x \, dz \, dy \, dx\]Each integration step narrows down the volume components and adjusts the overall outcome based on the bounds defined, thus allowing us to evaluate the total flux through the region.

Flux Calculation

Flux through a surface quantifies the amount of "stuff" (like fluid, air, or electricity) passing through that surface. In vector calculus, the concept of flux is crucial for understanding how vector fields interact with surfaces, which is where the Divergence Theorem comes in.
The Divergence Theorem relates the flux through a closed surface to the volume integral of a divergence inside that surface. Mathematically, it is expressed as \[ \iint_S (\mathbf{F} \cdot \mathbf{n}) \, dS = \iiint_D (abla \cdot \mathbf{F}) \, dV \]
The left-hand side denotes the surface integral of the field over the surface \(S\), where \(\mathbf{n}\) represents the outward normal vector. The right-hand side is the volume integral over the divergence of the field.
In the exercise, we simplify the calculation using symmetry and properties of the integral:\[ \int_{-3}^{3} \int_{x^2}^{9} \int_{0}^{9-y} 2x \, dz \, dy \, dx = 0 \] The result is obtained through detailed evaluation of each integral step, recognizing that certain types of functions (i.e., odd functions) over symmetric bounds yield zero.

• Integrals like \(\int_{-3}^{3} 81x \, dx\) vanish due to symmetry.
• Terms evaluated across symmetric limits also contribute zero due to offsetting negative and positive areas under the function curve.

This application of flux calculation through the Divergence Theorem illustrates the elegant balance between surface and volume measures in mathematics.