Problem 34 - 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. \(\mathbf{F}=e^{y} \mathbf{i}+e^{x} \mathbf{j}+18 y \mathbf{k} ; S\) that portion of the plane \(x+y+z=6\) in the first octant

Step-by-Step Solution

Answer

The flux is the result of evaluating the integrals in Step 8.

Step 1: Understand the Problem

We are asked to find the flux of the vector field \( \mathbf{F} = e^{y} \mathbf{i} + e^{x} \mathbf{j} + 18y \mathbf{k} \) through a surface \( S \). The surface \( S \) is part of the plane \( x+y+z=6 \) in the first octant, oriented upward.

Step 2: Determine the Surface Equation

The given plane equation is \( x + y + z = 6 \). In the first octant, \( x, y, \) and \( z \) must all be non-negative. The surface \( S \) can be described as the region \( x, y, z \geq 0 \), where \( z = 6 - x - y \).

Step 3: Find the Normal Vector

For a plane \( ax + by + cz = d \), the normal vector is \( \mathbf{n} = \langle a, b, c \rangle \). Thus, the normal vector to our plane is \( \mathbf{n} = \langle 1, 1, 1 \rangle \). We orient \( S \) upward, so we use \( \mathbf{n} = \langle 1, 1, -1 \rangle \) as the negative z-component is consistent with the upward orientation.

Step 4: Parameterize the Surface

Since \( z = 6 - x - y \), we parameterize \( S \) using parameters \( x \) and \( y \): \( \mathbf{r}(x, y) = \langle x, y, 6 - x - y \rangle \). The region of integration is the triangle in the \( xy \)-plane bounded by the x-axis, y-axis, and \( x+y=6 \).

Step 5: Compute the Flux Integral

The flux through \( S \) is given by \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} \). The differential area element is \( d\mathbf{S} = \mathbf{n} \, dA = \langle 1, 1, -1 \rangle \ \, dx \, dy \). The dot product \( \mathbf{F} \cdot \mathbf{n} = e^{y} + e^{x} - 18y \).

Step 6: Set Up the Integral

Calculate the flux: \[ \iint_{R} (e^{y} + e^{x} - 18y) \, dx \, dy \] where \( R \) is the region \( 0 \leq x \leq 6 \), \( 0 \leq y \leq 6-x \).

Step 7: Evaluate the Integral

Break the integral into separate parts and evaluate: \[\int_{0}^{6} \int_{0}^{6-x} e^{y} \, dy \, dx + \int_{0}^{6} \int_{0}^{6-x} e^{x} \, dy \, dx - \int_{0}^{6} \int_{0}^{6-x} 18y \, dy \, dx\]Calculate each part and add them to get the total flux.

Step 8: Calculate Each Part

Evaluate each integral component.1. \( \int_{0}^{6} \int_{0}^{6-x} e^{y} \, dy \, dx = \int_{0}^{6} [e^{y}]_{0}^{6-x} \, dx = \int_{0}^{6} (e^{6-x} - 1) \, dx \)2. \( \int_{0}^{6} \int_{0}^{6-x} e^{x} \, dy \, dx = \int_{0}^{6} e^{x}(6-x) \, dx \)3. \( \int_{0}^{6} \int_{0}^{6-x} 18y \, dy \, dx = \int_{0}^{6} [9y^2]_{0}^{6-x} \, dx \)

Key Concepts

Surface Integral

A surface integral is essentially an extension of a line integral. Instead of integrating over a curve, you calculate the integral over a surface. For vector fields, a surface integral helps you measure the flow of a field through a given surface.

When you calculate the surface integral of a vector field, you're essentially summing the flux through small surface elements. Flux is the quantity of the field passing through a surface. Mathematically, for a vector field \( \mathbf{F} \), the surface integral is given by \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} \).

• \( \mathbf{F} \) is the vector field.
• \( S \) is the surface you're evaluating on.
• \( \cdot \) denotes the dot product, combining field and normal vectors to find flow through \( S \).
• \( d\mathbf{S} \) is a vector normal to \( S \) representing a tiny surface segment.

This integral calculates how much of the field flows through \( S \), taking into account the surface orientation.

Parameterization

Parameterization involves expressing a surface using variables or parameters. It's crucial when you're dealing with complex surfaces. This method helps in transforming surface integration into a more manageable form. For a surface given in the form \( z = f(x,y) \), parameterization can be \( \mathbf{r}(x, y) = \langle x, y, f(x,y) \rangle \).

In this exercise, the surface equation is \( x + y + z = 6 \). Solving for \( z \), we get \( z = 6 - x - y \). Therefore, the vector function parameterizing the surface becomes \( \mathbf{r}(x, y) = \langle x, y, 6 - x - y \rangle \).

• This parameterization simplifies the plane into a form easier for integration.
• Using \( x \) and \( y \) as parameters, each point on the surface is described in terms of these variables.

Parameterization reduces the three-dimensional problem to two dimensions, allowing for straightforward computation of the surface integral.

Normal Vector Calculation

The normal vector is vital for determining the flow direction through a surface. It is perpendicular to the surface at a point and is used in the dot product calculation for the surface integral. The formula for the normal vector to a plane \( ax + by + cz = d \) is \( \mathbf{n} = \langle a, b, c \rangle \).

For the plane \( x + y + z = 6 \), the normal vector initially is \( \mathbf{n} = \langle 1, 1, 1 \rangle \). However, to orient the surface upward, we modify this to \( \mathbf{n} = \langle 1, 1, -1 \rangle \).

• Direction matters: Changing the sign of a component alters the orientation.
• For upward orientation, the z-component reflects the outward direction in the positive z-axis context.

Calculating the normal vector ensures the surface is properly aligned with the vector field, directing the flux computation accordingly.

Triangle Region in xy-plane

In many exercises involving surface integrals, you parameterize over a specific region. Here, the triangular region is crucial as it's where the parameters \( x \) and \( y \) vary. The triangle is enclosed in the xy-plane, bounded by the x-axis, y-axis, and line \( x+y = 6 \).

This triangular region determines limits for integration functions over \( x \) and \( y \).

• x-axis: \( y = 0 \)
• y-axis: \( x = 0 \)
• Line \( x+y=6 \): delineates the maximum region a point can reach on this surface

The limits go from \( 0 \leq x \leq 6 \) and \( 0 \leq y \leq 6-x \). Each of these boundaries forms a straight edge of the triangle within the plane.
Reminding yourself of these boundaries ensures integrating in the correct region for calculations.