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Magazine Chapter 19: Problem 30
Compute the flux of the vector field \(\vec{F}\) through the surface \(S\). \(\vec{F}=z \vec{k}\) and \(S\) is the portion of the plane \(x+y+z=1\) that lies in the first octant, oriented upward.
Short Answer
Expert verified
The flux through the surface is \( \frac{1}{6} \).
Step by step solution
01
Understand the Problem
We are tasked with calculating the flux of the vector field \( \vec{F} = z \vec{k} \) through a specific surface \( S \). The surface \( S \) is the part of the plane \( x + y + z = 1 \) in the first octant (where \( x, y, z \geq 0 \)). The surface is oriented upward, meaning the normal vector will point in the positive \( z \)-direction.
02
Determine Parameterization of the Surface
The plane can be described by \( x + y + z = 1 \). In the first octant, the limits for \( x \) and \( y \) are 0 to 1 and 0 to \( 1-x \) respectively. We parameterize \( z \) in terms of \( x \) and \( y \) as \( z = 1 - x - y \). Thus, the parameterization of the surface is \( \vec{r}(x, y) = \langle x, y, 1 - x - y \rangle \).
03
Find the Normal Vector
To find the normal vector, compute the cross product of the partial derivatives \( \vec{r}_x \) and \( \vec{r}_y \). \( \vec{r}_x = \langle 1, 0, -1 \rangle \) and \( \vec{r}_y = \langle 0, 1, -1 \rangle \). The cross product \( \vec{n} = \vec{r}_x \times \vec{r}_y = \langle 1, 1, 1 \rangle \) gives a normal vector pointing upward as required by the problem.
04
Set Up the Surface Integral
The surface integral for flux is given by \( \iint_S \vec{F} \cdot \vec{n} \, dS \). Substituting \( \vec{F} = z \vec{k} = (1 - x - y) \) and \( \vec{n} = \langle 1, 1, 1 \rangle \), we have \( \iint_S (0 + 0 + (1-x-y)) \cdot 1 \, dS = \iint_D (1-x-y) \, dA \), where \( D \) is the region \( x \geq 0, y \geq 0, x + y \leq 1 \).
05
Evaluate the Double Integral
The region \( D \) is a right triangle with vertices \( (0,0), (1,0), (0,1) \). The integral becomes \( \int_0^1 \int_0^{1-x} (1-x-y) \, dy \, dx \). Evaluate the inner integral: \( \int_0^{1-x} (1-x-y) \, dy = [(1-x)y - \frac{y^2}{2}]_0^{1-x} = (1-x)(1-x) - \frac{(1-x)^2}{2} = \frac{(1-x)^2}{2} \). Now, evaluate the outer integral: \( \int_0^1 \frac{(1-x)^2}{2} \, dx = \left[ -\frac{1}{6}(1-x)^3 \right]_0^1 = \frac{1}{6} \).
06
Conclusion: Calculating the Flux
The value of the flux of the vector field \( \vec{F} \) through the surface \( S \) is \( \frac{1}{6} \). This completes our solution to the problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Surface Integral
A surface integral allows us to measure how a vector field interacts with a surface. In this context, it's about determining the flux of a vector field across a surface in three-dimensional space. The
- Vector Field: A mathematical function assigning a vector to each point in space. Here, we use \(\vec{F} = z \vec{k}\), meaning the field influences only along the z-axis.
- Surface: In this problem, this is the section of the plane \(x + y + z = 1\) existing in the first octant.
- \(\vec{n}\): Represents a normal vector perpendicular to the surface.
- \(dS\): A small element of the surface area.
Parameterization
Parameterization is the process of expressing a surface using two parameters, allowing us to work with complex shapes in calculus. In our problem, the plane \(x+y+z=1\) needs to be characterized in terms of parameters \(x\) and \(y\). Here's how that happens:
- Surface Equation: The plane is \(x+y+z=1\).
- First Octant Restrictions: Since we are in the first octant, \(x, y,\) and \(z\) are non-negative.
- Express \(z\): We write \(z = 1 - x - y\).
Normal Vector
A normal vector is crucial when calculating a surface integral, as it defines the orientation of the surface. Here's how to find it for this problem:
- Parameter Derivatives: We need the surface's partial derivatives, \(\vec{r}_x\) and \(\vec{r}_y\).
- Calculating Partials: The derivatives are \(\vec{r}_x = \langle 1, 0, -1 \rangle\) and \(\vec{r}_y = \langle 0, 1, -1 \rangle\).
First Octant
In three-dimensional algebraic geometry, the first octant is a vital concept. It describes a part of space where coordinates of points are all non-negative:
- Definition: It is defined by \(x \geq 0\), \(y \geq 0\), and \(z \geq 0\).
- Conceptualization: Imagine octants like parts of a Cartesian coordinate system divided by planes intersecting at the origin.
- Only a portion of the entire plane \(x + y + z = 1\) is considered.
- It defines clear integration limits for the problem, trapping \(x\) and \(y\) between \(0\) and \(1 - x\).
