Problem 29 - Advanced Engineering Mathematics

Question

Let \(\mathbf{F}\) be a vector field. Find the flux of \(\mathbf{F}\) through the given surface. Assume the surface \(S\) is oriented upward. \(\mathbf{F}=x \mathbf{i}+2 z \mathbf{j}+y \mathbf{k} ; S\) that portion of the cylinder \(y^{2}+z^{2}=4\) in the first octant bounded by \(x=0, x=3, y=0, z=0\)

Step-by-Step Solution

Answer

The flux through the surface is 12.

Step 1: Understand the Problem

We need to calculate the flux of the vector field \(\mathbf{F} = x \mathbf{i} + 2z \mathbf{j} + y \mathbf{k}\) over the surface of the cylinder defined by \(y^2 + z^2 = 4\). The cylinder is in the first octant and bounded by the planes \(x=0\), \(x=3\), \(y=0\), and \(z=0\). The orientation of the surface is upward.

Step 2: Parameterize the Surface

The given surface is a cylinder. We can use cylindrical coordinates to parameterize it. Let \(y = 2\sin\theta\) and \(z = 2\cos\theta\) where \(0 \leq \theta \leq \frac{\pi}{2}\). Since the surface is bounded by \(x=0\) and \(x=3\), we have the parameters \(0 \leq x \leq 3\). Thus, the parameterization is: \(\mathbf{r}(x, \theta) = x \mathbf{i} + 2\sin\theta \mathbf{j} + 2\cos\theta \mathbf{k}\).

Step 3: Find the Normal Vector

To find the flux, we need the normal vector to the surface. The normal vector is given by the cross product of the partial derivatives of \(\mathbf{r}\) with respect to \(x\) and \(\theta\). Compute \(\frac{\partial \mathbf{r}}{\partial x} = \mathbf{i}\) and \(\frac{\partial \mathbf{r}}{\partial \theta} = 2\cos\theta \mathbf{j} - 2\sin\theta \mathbf{k}\). The cross product is \(\mathbf{N} = \mathbf{i} \times (2\cos\theta \mathbf{j} - 2\sin\theta \mathbf{k}) = 2\sin\theta \mathbf{j} + 2\cos\theta \mathbf{k}\).

Step 4: Calculate the Flux Integral

The flux \(\Phi\) through the surface \(S\) is given by \(\iint_{S} \mathbf{F} \cdot \mathbf{N} \ dS\). Here, \(\mathbf{F} \cdot \mathbf{N} = (x \mathbf{i} + 2z \mathbf{j} + y \mathbf{k}) \cdot (2\sin\theta \mathbf{j} + 2\cos\theta \mathbf{k}) \). Substitute \(y = 2\sin\theta\) and \(z = 2\cos\theta\) to get: \([2z \cdot 2\sin\theta + y \cdot 2\cos\theta] = [4\cos\theta \sin\theta + 4\sin\theta \cos\theta] = 8 \cos\theta \sin\theta\).

Step 5: Evaluate the Double Integral

The surface integral becomes \(\int_{0}^{3} \int_{0}^{\frac{\pi}{2}} 8 \cos\theta \sin\theta \ d\theta \ dx\). Evaluate the inner integral \(\int_{0}^{\frac{\pi}{2}} 8 \cos\theta \sin\theta \ d\theta\). This can be simplified using the identity \(\sin 2\theta = 2\sin\theta \cos\theta\), thus the integral becomes \(4\int_{0}^{\frac{\pi}{2}} \sin 2\theta \ d\theta\). Integrate to get \([-2\cos 2\theta]_{0}^{\frac{\pi}{2}} = 2 + 2 = 4\). Now integrate over \(x\): \(\int_{0}^{3} 4 \ dx = 12\).

Step 6: Conclusion and Result

The calculated flux of the vector field \(\mathbf{F}\) through the surface of the cylinder is 12.

Key Concepts

Vector field

A vector field is a mathematical construct that assigns a vector to every point in space. In physics and engineering, vector fields often represent the distribution of some quantity, like force or velocity, in a region of space. Here, we're dealing with the vector field \[ \mathbf{F} = x \mathbf{i} + 2z \mathbf{j} + y \mathbf{k} \] where the vector components are functions of spatial coordinates \(x\), \(y\), and \(z\). This field can be thought of as describing how a certain physical quantity (such as a fluid or a force field) changes across the three-dimensional space.

• \(x \mathbf{i}\) indicates the influence along the x-direction.
• \(2z \mathbf{j}\) indicates the influence along the y-direction, dependent on \(z\).
• \(y \mathbf{k}\) indicates the influence along the z-direction, dependent on \(y\).

Understanding a vector field like \(\mathbf{F}\) is crucial when calculating flux as it represents how the field interacts with surfaces in space, like our cylinder.

Cylinder surface integration

Surface integration over a cylinder involves calculating how much a vector field passes perpendicularly through a surface. This is crucial in physical contexts, such as determining the amount of fluid going through a filtering surface. To find the flux through the cylinder's surface bounded in the first octant, we need to set up and solve an integral over that surface.The surface is defined by the cylindrical equation \(y^2 + z^2 = 4\), and the integration involves evaluating the cross product of surface parameters to find a normal vector. This normal vector, \[ \mathbf{N} = 2\sin\theta \mathbf{j} + 2\cos\theta \mathbf{k} \] is essential to determine how the vector field \(\mathbf{F}\) interacts perpendicularly with the surface.
By parameterizing the surface accordingly and correctly setting the limits for integration, we can efficiently compute the flux integral, which tells us the total flow across the surface.

Cylindrical coordinates

Cylindrical coordinates are vital when dealing with problems involving cylindrical shapes, providing a more natural description than Cartesian coordinates. In this system, the position of a point in space is described by:

• \(r\): the radial distance from the z-axis.
• \(\theta\): the angle from the x-axis.
• \(z\): the height above the xy-plane.

For our cylindrical surface \(y^2 + z^2 = 4\), we use these coordinates by letting \(y = 2\sin\theta\) and \(z = 2\cos\theta\), a transformation that simplifies our calculations. These parameters map perfectly onto the circular cross-sections of our cylinder.
Using this, we simplify the expression of our vector field and the normal vector for subsequent integration, making our task of computing flux integrals much more manageable.

Double integrals

The use of double integrals is crucial in evaluating the flux through a surface. A double integral allows us to sum over a two-dimensional area, accounting for variations across both dimensions of the surface. For the given problem, we must solve a double integral of the form: \[\int_{0}^{3} \int_{0}^{\frac{\pi}{2}} (\text{Expression in } \theta) \ d\theta \ dx \]The expression inside typically represents the dot product of the vector field \(\mathbf{F}\) with the normal vector \(\mathbf{N}\), which gives us the magnitude of the field flowing through each differential element of the surface. By solving the inner integral first (holding x constant) and then the outer one, we effectively cover the entire surface.
This two-stage integration is a powerful tool to compute quantities over complex surfaces and yields the total flux, a single scalar quantity, summarizing the field's interaction with the surface. Understanding how to set up and evaluate these integrals is key to mastering flux calculations.