Chapter 6: Problem 302
\(\quad\) Compute \(\quad \iint_{S} \mathbf{F} \cdot \mathbf{N} d S,\) where \(\mathbf{F}(x, y, z)=x \mathbf{i}-5 y \mathbf{j}+4 z \mathbf{k}\) and \(\mathbf{N}\) is an outward normal vector \(S\), where \(S\) is the union of two squares \(S_{1}: x=0,0 \leq y \leq 1,0 \leq z \leq 1\) and \(S_{2}: z=1,0 \leq x \leq 1,0 \leq y \leq 1\)
Short Answer
Step by step solution
Understand the Problem
Determine Normal Vector for \( S_1 \)
Set up the integral for \( S_1 \)
Determine Normal Vector for \( S_2 \)
Set up the integral for \( S_2 \)
Calculate the Integral for \( S_2 \)
Combine Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Field Integration
When performing vector field integration, we consider a surface integral of the form \(\iint_{S} \mathbf{F} \cdot \mathbf{N} \, dS\). Here, \(\mathbf{F}\) is a vector field, \(\mathbf{N}\) is the normal vector to the surface, and \(S\) is the surface over which the integration occurs.
- Dot Product: We take the dot product of the vector field \(\mathbf{F}\) and the outward normal vector \(\mathbf{N}\). This helps us determine the component of the vector field that is perpendicular to the surface.
- Surface Element \(dS\): This represents a small piece of the surface, allowing integration to occur over the entire surface area.
Outward Normal Vector
In our problem, we have two distinct surfaces, each with a unique outward normal vector.
- Surface \(S_1\): For the square defined by \(x = 0\), the outward normal vector \(\mathbf{N}\) is \(-\mathbf{i}\). This points along the negative x-direction since \(x\) is constant.
- Surface \(S_2\): For the square defined at \(z = 1\), the normal vector is \(\mathbf{k}\). This points in the positive z-direction as it is situated at the plane where \(z = 1\).
Surface Orientation
For the exercise at hand, the surface \(S\) is divided into two parts:
- Square \(S_1\): The surface lies on the plane where \(x = 0\). Given this boundary, we orient \(S_1\) ensuring the normal vector is negative with respect to the axis. Hence, the outward orientation results in \(-\mathbf{i}\).
- Square \(S_2\): This component is positioned at \(z = 1\) with an obvious upward orientation. Here, \(\mathbf{k}\) is naturally the outward normal vector. Creating an orientation that aligns with the increasing z-axis.