Problem 302 - Calculus Volume 3

Question

\(\quad\) 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 \(S_{2}: z=1,0 \leq x \leq 1,0 \leq y \leq 1\)

Step-by-Step Solution

Answer

The surface integral over \( S \) is 4.

Step 1: Understand the Problem

We need to compute the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{N} \, dS \) over the surface \( S \), which is composed of two parts: \( S_1 \) and \( S_2 \). Each part has a different orientation and needs to be treated separately.

Step 2: Determine Normal Vector for \( S_1 \)

For the square \( S_1 \) defined by \( x = 0 \), the outward normal vector is \( \mathbf{N} = -\mathbf{i} \). This is because the surface is in the plane where \( x = 0 \) and points outward.

Step 3: Set up the integral for \( S_1 \)

The surface integral over \( S_1 \) becomes \( \iint_{S_1} \mathbf{F} \cdot \mathbf{N} \, dS = \iint_{0 \leq y \leq 1, 0 \leq z \leq 1} (x, -5y, 4z) \cdot (-1, 0, 0) \, dydz \). This simplifies to \( \iint_{0 \leq y \leq 1, 0 \leq z \leq 1} (-x) \, dydz \). Since \( x = 0 \) on \( S_1 \), the integral becomes \( 0 \).

Step 4: Determine Normal Vector for \( S_2 \)

For the square \( S_2 \) defined by \( z = 1 \), the outward normal vector is \( \mathbf{N} = \mathbf{k} \). This orientation is outward from the surface, matching the given direction.

Step 5: Set up the integral for \( S_2 \)

The surface integral over \( S_2 \) is \( \iint_{S_2} \mathbf{F} \cdot \mathbf{N} \, dS = \iint_{0 \leq x \leq 1, 0 \leq y \leq 1} (x, -5y, 4z) \cdot (0, 0, 1) \, dxdy \). This simplifies to \( \iint_{0 \leq x \leq 1, 0 \leq y \leq 1} 4z \, dxdy \). Since \( z = 1 \) on \( S_2 \), the integral becomes \( \iint_{0 \leq x \leq 1, 0 \leq y \leq 1} 4 \, dxdy \).

Step 6: Calculate the Integral for \( S_2 \)

Calculate the integral: \[ \int_0^1 \int_0^1 4 \, dx \, dy = \int_0^1 \left[ 4x \right]_0^1 \, dy = \int_0^1 4 \, dy = \left[ 4y \right]_0^1 = 4 \].

Step 7: Combine Results

Add the results from \( S_1 \) and \( S_2 \). From Steps 3 and 6, the result is \( 0 + 4 = 4 \).

Key Concepts

Vector Field Integration

Vector field integration is a process where we compute the integral of a vector field over a given surface. It is a crucial concept in fields like fluid dynamics and electromagnetism. By integrating a vector field, we can determine quantities like the flow of a fluid across a surface or the flux of an electric field through a given area.

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.

To solve the example problem, we first identify the separate parts of the surface \(S\) and proceed to compute the integrals for each part. Each integral considers the direction and magnitude of the vector field relative to the surface orientation.

Outward Normal Vector

The outward normal vector is paramount in evaluating surface integrals. This vector is perpendicular to the surface and points outward, away from the region being considered. Knowing the correct direction of the normal vector is essential because it informs us of the correct orientation of the surface.

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 \(z = 1\).

The choice of an outward normal vector depends heavily on the specific boundaries and orientation of the surface provided in a problem. By carefully considering the geometric and physical context, we correctly align our calculations with the intended vector field mapping.

Surface Orientation

Surface orientation plays a central role in the calculation of surface integrals. It determines how we align our calculations and integrate the vector field across \(S\). Understanding this orientation involves knowing how \(S\) is composed and the direction of the normals.

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.

Correctly establishing the orientation guides the direction of the outward normal vectors. It helps ensure each part of the integral aligns in capturing the right portion of the vector field as dictated by these orientations. This entire process ensures the correct calculation of the involved vector field in relation to its spatial setting.