Problem 14 - 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 \(\mathbf{F}\). $$ \begin{aligned} &\mathbf{F}=x y^{2} \mathbf{i}+x^{2} y \mathbf{j}+6 \sin x \mathbf{k} ; D \text { the region bounded by the }\\\ &\text { cone } z=\sqrt{x^{2}+y^{2}} \text { and the planes } z=2, z=4 \end{aligned} $$

Step-by-Step Solution

Answer

The outward flux is \( 480\pi \).

Step 1: Identify the Problem

We are tasked with finding the outward flux of the vector field \( \mathbf{F} = x y^{2} \mathbf{i} + x^{2} y \mathbf{j} + 6 \sin x \mathbf{k} \) across the surface \( S \). \( S \) is the boundary of the region \( D \) enclosed by the cone \( z = \sqrt{x^2 + y^2} \) and the planes \( z = 2 \) and \( z = 4 \).

Step 2: Apply the Divergence Theorem

The divergence theorem states \( \iint_{S} (\mathbf{F} \cdot \mathbf{n}) \, dS = \iiint_{D} (abla \cdot \mathbf{F}) \, dV \). We need to calculate the divergence \( abla \cdot \mathbf{F} \) and integrate over the region \( D \).

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

Calculate \( abla \cdot \mathbf{F} \) where \( \mathbf{F} = (xy^2, x^2y, 6\sin x) \). The divergence is: \[abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xy^2) + \frac{\partial}{\partial y}(x^2y) + \frac{\partial}{\partial z}(6\sin x).\]Thus, \( abla \cdot \mathbf{F} = y^2 + x^2 + 0 = y^2 + x^2 \).

Step 4: Set Up the Integral

The divergence of the vector field is \( y^2 + x^2 \). We will integrate this over the volume \( D \), where \( D \) is bounded by \( 2 \leq z \leq 4 \) and \( z = \sqrt{x^2 + y^2} \) defines a cone.The transformation to cylindrical coordinates simplifies the integration:- \( x = r \cos \theta \)- \( y = r \sin \theta \)- \( z = r \)Thus, \( 2 \leq z \leq 4 \) translates to \( 2 \leq r \leq 4 \).

Step 5: Evaluate the Triple Integral

Substitute the expressions and limits into the integral in cylindrical coordinates:\[\iiint_{D} (y^2 + x^2) \, dV = \int_{0}^{2\pi} \int_{0}^{4} \int_{2}^{4} (r^2 \sin^2 \theta + r^2 \cos^2 \theta) r \, dz \, dr \, d\theta.\]This simplifies to:\[\int_{0}^{2\pi} \int_{0}^{4} \int_{2}^{4} r^3 \, dz \, dr \, d\theta.\]

Step 6: Integrate with Respect to \( z \)

The integration with respect to \( z \) from 2 to 4 gives:\[\int_{2}^{4} r^3 \, dz = r^3 (z \big|_{2}^{4}) = r^3 (4-2) = 2r^3.\]

Step 7: Integrate with Respect to \( r \) and \( \theta \)

Perform the integration with respect to \( r \) from 2 to 4:\[\int_{2}^{4} 2r^3 \, dr = 2 \left[ \frac{r^4}{4} \right]_2^4 = 2 \left( \frac{256}{4} - \frac{16}{4} \right) = 240.\]Finally, integrate with respect to \( \theta \) from 0 to \( 2\pi \):\[\int_{0}^{2\pi} 240 \, d\theta = 240 (2\pi) = 480\pi.\]

Step 8: Final Result

The outward flux of the vector field \( \mathbf{F} \) across the surface \( S \) is \( 480\pi \).

Key Concepts

Vector Field

A vector field is essentially a function that assigns a vector to every point in space. Imagine the wind flowing through the air where each point in the atmosphere has a vector that represents the speed and direction of the wind at that point. Vector fields can be represented in three dimensions using components along the x, y, and z axes, typically written as \(\mathbf{F}(x, y, z) = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k}\). Here, \(P, Q, R\) are functions of \(x, y, z\).

For the given exercise, the vector field is \(\mathbf{F} = xy^2 \mathbf{i} + x^2y \mathbf{j} + 6 \sin x \mathbf{k}\). Each component (\(xy^2, x^2y, 6 \sin x\)) represents influences or behaviors in different directions at any point \((x, y, z)\). Understanding vector fields helps us analyze physical phenomena like fluid flow or electromagnetic fields.

Outward Flux

Outward flux is the measure of how much of the vector field passes outwardly through a given surface. It is crucial in fields such as fluid dynamics where it might represent the amount of fluid flowing out of a surface.

To calculate outward flux, we use the formula \(\iint_{S}(\mathbf{F} \cdot \mathbf{n}) \, d S\), where \(\mathbf{F}\) is the vector field and \(\mathbf{n}\) is the unit normal to the surface \(S\). This formula effectively calculates how the vector field "cuts" across the surface. In the provided problem, this involves using the divergence theorem to transform the surface integral into a volume integral over region \(D\) which calculates total divergence within \(D\), making computation simpler.

Cylindrical Coordinates

Cylindrical coordinates are a coordinate system that extends polar coordinates by adding a height dimension. They are particularly useful in problems involving rotational symmetries around an axis.

Cylindrical coordinates \((r, \theta, z)\) relate to Cartesian coordinates \((x, y, z)\) as follows:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(z = z\)

This transformation simplifies integration in regions like our problem involving the cone \(z = \sqrt{x^2 + y^2}\). In the solution, the set of equations allows conversion into cylindrical coordinates, reducing the complexity of evaluating the triple integral and accommodating the symmetry of the cone's shape.

Triple Integral

Triple integrals are an extension of the concept of integration into three dimensions. They allow us to compute volume integrals, which are useful for calculating quantities in three-dimensional spaces.

In the specific exercise, the triple integral \[\iiint_{D} (y^2 + x^2) \, dV\] is set up using cylindrical coordinates. Each element \(dV\) represents an infinitesimally small volume element in the region \(D\), where the triple integral evaluates the function \(y^2 + x^2\) over the entire region enclosed by the cone and planes. This involves integrating over \(z\), then \(r\), and finally \(\theta\). The limits of integration define the extent of the cone and planes, transforming the integral from a volume to a solvable mathematical expression that yields \(480\pi\) as the outward flux.