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Magazine Chapter 16: Problem 23
Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Plane } \mathbf{F}=2 x y \mathbf{i}+2 y z \mathbf{j}+2 x z \mathbf{k} \text { upward across the portion of }} \\ {\text { the plane } x+y+z=2 a \text { that lies above the square } 0 \leq x \leq a,} \\\ {0 \leq y \leq a, \text { in the } x y-\text { plane }}\end{array}\)
Short Answer
Expert verified
The flux across the surface is \( \frac{8a^3}{3} \).
Step by step solution
01
Understand the Problem
We need to find the flux of vector field \( \mathbf{F} = 2xy\mathbf{i} + 2yz\mathbf{j} + 2xz\mathbf{k} \) across a plane given by \( x+y+z=2a \). We are considering only the portion of this plane that lies above the square region \( 0 \leq x \leq a \) and \( 0 \leq y \leq a \), and need to calculate the flux in the upward direction.
02
Determine the Surface Parametrization
To parametrize the surface, express \( z \) in terms of \( x \) and \( y \) using the plane equation \( x + y + z = 2a \), leading to \( z = 2a - x - y \). The parametrization \( \mathbf{r}(x,y) = \langle x, y, 2a-x-y \rangle \) fits within the square region.
03
Find the Normal Vector
The normal vector to the parametric surface is found using cross products of partial derivatives. Compute:\[ \frac{\partial \mathbf{r}}{\partial x} = \langle 1, 0, -1 \rangle \] and \[ \frac{\partial \mathbf{r}}{\partial y} = \langle 0, 1, -1 \rangle \].The cross product \( \mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial y} = \langle 1, 1, 1 \rangle \).
04
Dot Product with Vector Field
Calculate the dot product \( \mathbf{F} \cdot \mathbf{n} = (2xy)1 + (2yz)1 + (2xz)1 = 2xy + 2y(2a-x-y) + 2x(2a-x-y) \). Simplify to \( 4ax + 4ay - 4xy - y^2 - x^2 \).
05
Setup and Solve the Double Integral
Calculate the flux integral \[ \iint_{0}^{a} \iint_{0}^{a} (4ax + 4ay - 4xy - y^2 - x^2) \: dy \, dx \].Integrate with respect to \( y \) first and then \( x \). Simplify to find constants and bounds.
06
Evaluate the Integral
Computing each term's integral separately, evaluate:- \( \int_{0}^{a} \int_{0}^{a} 4ax \, dy \, dx = 2a^3 \)- \( \int_{0}^{a} \int_{0}^{a} 4ay \, dy \, dx = 2a^3 \)- \( \int_{0}^{a} \int_{0}^{a} 4xy \, dy \, dx = a^3 \)- \( \int_{0}^{a} \int_{0}^{a} y^2 \, dy \, dx = \frac{a^3}{3} \)- \( \int_{0}^{a} \int_{0}^{a} x^2 \, dy \, dx = \frac{a^3}{3} \)Summing these results gives the total flux across the surface.
07
Final Step: Solution Summation
Combine the evaluated integral results: \[ 2a^3 + 2a^3 - a^3 - \frac{a^3}{3} - \frac{a^3}{3} \] equals \( \frac{8a^3}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Field
A vector field is a mathematical construct where each point in space is associated with a vector. Here, our vector field is defined as \( \mathbf{F} = 2xy \mathbf{i} + 2yz \mathbf{j} + 2xz \mathbf{k} \). This tells us that at any point \((x, y, z)\) in the space, there's a vector with components that depend on these coordinates.
Each component of \( \mathbf{F} \) represents the influence of its respective axis-direction on the field. Specifically:
Each component of \( \mathbf{F} \) represents the influence of its respective axis-direction on the field. Specifically:
- \(2xy\mathbf{i}\) indicates the influence in the x-direction, scaling with both the x and y coordinates.
- \(2yz\mathbf{j}\) reflects changes in the y-direction, with influence from the y and z coordinates.
- \(2xz\mathbf{k}\) represents the effect in the z-direction, depending on the x and z coordinates.
Surface Parametrization
Surface parametrization involves expressing surfaces using parameters rather than explicit equations. In this case, we have the plane \( x + y + z = 2a \).
To parametrize this surface, we express \( z \) in terms of \( x \) and \( y \): \( z = 2a - x - y \). This lets us define a parametric representation \( \mathbf{r}(x, y) = \langle x, y, 2a-x-y \rangle \), covering every point on the plane for \( 0 \leq x, y \leq a \).
Surface parametrization is powerful because it transforms complex surface equations into parameter-driven functions, simplifying further calculations like normal vectors, essential for flux calculations.
To parametrize this surface, we express \( z \) in terms of \( x \) and \( y \): \( z = 2a - x - y \). This lets us define a parametric representation \( \mathbf{r}(x, y) = \langle x, y, 2a-x-y \rangle \), covering every point on the plane for \( 0 \leq x, y \leq a \).
Surface parametrization is powerful because it transforms complex surface equations into parameter-driven functions, simplifying further calculations like normal vectors, essential for flux calculations.
Flux Calculation
Flux measures how much of a vector field passes through a given surface. It gives us a sense of the flow across the surface. Here, it's computed as the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \), where \( \mathbf{n} \) is the normal vector.
First, we derive the normal vector to the surface using the cross product of partial derivatives from the parameterization.
For this surface:
The cross product \( \mathbf{n} = \langle 1, 1, 1 \rangle \), which is the normal to the surface pointing in the upward direction. The dot product \( \mathbf{F} \cdot \mathbf{n} \) = \( 2xy + 2yz + 2xz \) is then calculated to evaluate the flux through this surface.
First, we derive the normal vector to the surface using the cross product of partial derivatives from the parameterization.
For this surface:
- The partial derivative with respect to \( x \) is \( \langle 1, 0, -1 \rangle \).
- The partial derivative with respect to \( y \) is \( \langle 0, 1, -1 \rangle \).
The cross product \( \mathbf{n} = \langle 1, 1, 1 \rangle \), which is the normal to the surface pointing in the upward direction. The dot product \( \mathbf{F} \cdot \mathbf{n} \) = \( 2xy + 2yz + 2xz \) is then calculated to evaluate the flux through this surface.
Double Integral Evaluation
Double integrals allow us to compute quantities over a two-dimensional region. In this exercise, the double integral evaluates the flux across a surface defined in \((x, y)\)-plane.
We setup the integral as \( \iint (4ax + 4ay - 4xy - y^2 - x^2) \, dy \, dx\), representing the dot product of the vector field and the normal vector over the square region \(0 \leq x, y \leq a\).
This setup involves breaking the integral into parts and evaluating each component:
We setup the integral as \( \iint (4ax + 4ay - 4xy - y^2 - x^2) \, dy \, dx\), representing the dot product of the vector field and the normal vector over the square region \(0 \leq x, y \leq a\).
This setup involves breaking the integral into parts and evaluating each component:
- Integrate each term with respect to \( y \) then \( x \).
- Compute the individual integral results, such as \( \int_{0}^{a} \int_{0}^{a} 4ax \, dy \, dx = 2a^3 \) and others similarly.
- Sum these results: \(2a^3 + 2a^3 - a^3 - \frac{a^3}{3} - \frac{a^3}{3}\).
