Problem 29 - Thomas Calculus

Question

Find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. \(\begin{array}{l}{\mathbf{F}(x, y, z)=-\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}} \\\ {S : \text { rectangular surface } z=0, \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 3} \\ {\quad \text { direction } \mathbf{k}}\end{array}\)

Step-by-Step Solution

Answer

The surface integral is 18.

Step 1: Identify the Type of Integral and Surface

We need to compute the surface integral \(\iint_S \mathbf{F} \cdot d\mathbf{S}\), where \(S\) is a plane within the XY-plane, specifically \(z=0\) with bounds \(0 \leq x \leq 2\) and \(0 \leq y \leq 3\). The direction is given as \(\mathbf{k}\), indicating we are dealing with an upward orientation normal to the XY plane.

Step 2: Determine the Surface's Normal Vector

Since the surface \(S\) lies on the plane \(z=0\), the unit normal vector \(\mathbf{n}\) is in the direction of \(\mathbf{k}\), which can be represented as \(\mathbf{n} = \mathbf{k} = \langle 0, 0, 1 \rangle\). This vector indicates the surface is oriented upward.

Step 3: Calculate \(\mathbf{F} \cdot \mathbf{n}\)

Substitute \(\mathbf{n}\) into the dot product \(\mathbf{F} \cdot \mathbf{n}\). Here, \(\mathbf{F}(x, y, z) = \langle -1, 2, 3 \rangle\) and \(\mathbf{n} = \langle 0, 0, 1 \rangle\), so:\[\mathbf{F} \cdot \mathbf{n} = \langle -1, 2, 3 \rangle \cdot \langle 0, 0, 1 \rangle = -1\cdot0 + 2\cdot0 + 3\cdot1 = 3\]

Step 4: Set Up the Surface Integral

The surface integral over \(S\) is given by:\[\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S (3) \, dS\]Where \(dS\) is the differential area element in the XY-plane for the surface at \(z=0\), translated to \(dx \, dy\).

Step 5: Evaluate the Double Integral

Express the double integral over the region \(S\):\[\int_{0}^{3} \int_{0}^{2} 3 \, dx \, dy\]First, evaluate the inner integral with respect to \(x\):\[\int_{0}^{2} 3 \, dx = 3x \Big |_{0}^{2} = 3(2) - 3(0) = 6\]Next, evaluate the outer integral with respect to \(y\):\[\int_{0}^{3} 6 \, dy = 6y \Big |_{0}^{3} = 6(3) - 6(0) = 18\]

Step 6: Final Surface Integral Result

The surface integral evaluates to \(18\). Therefore, the integral of \(\mathbf{F}\) over the given surface in the specified direction is \(18\).

Key Concepts

Understanding Vector Field

A vector field is a mathematical representation where each point in space is associated with a vector. In simpler terms, it's like having arrows attached to every point in a region, showing the direction and magnitude of a quantity like force or velocity. In our problem, the vector field \( \mathbf{F}(x, y, z) = \langle -1, 2, 3 \rangle \) is constant, meaning every vector across the field has the same direction and length: one unit in the negative x-direction, two units in the positive y-direction, and three units in the positive z-direction. - Think of this as a breeze blowing through a field, where the wind has a consistent strength and direction everywhere.- In graphics or physics, vector fields help visualize how objects interact within a space, by indicating paths of flow like water streams or gravitational fields.These fields play a crucial role in calculating surface integrals, as they determine how the entire surface interacts with the vector field, impacting the final integral result.

Exploring the Double Integral

A double integral is a method used to calculate the volume under a surface defined over a two-dimensional area. It extends the concept of single-variable integration, which finds the area under a curve, to two dimensions, covering more than just lines and curves. In our scenario, \[ \iint_S (3) \, dS \] represents integrating a constant function, 3, over a region in the xy-plane. The region is defined by boundaries: 0 ≤ x ≤ 2 and 0 ≤ y ≤ 3.- First, the inner integral \( \int_{0}^{2} 3 \, dx = 6 \) calculates the area across x for a constant y value.- Then, the outer integral \( \int_{0}^{3} 6 \, dy = 18 \) aggregates results over the y-direction, summing up these volumes to give the final total.Using the double integral approach helps derive quantities like total mass, charge, or flow, across a defined surface or region in physical problems.

Importance of the Normal Vector

A normal vector is an essential tool in vector calculus. It is a vector that is perpendicular to a surface at a given point, which helps determine how that surface interacts with external forces or fields.In the exercise, the normal vector \( \mathbf{n} = \mathbf{k} = \langle 0, 0, 1 \rangle \) indicates that the surface is lying flat in the xy-plane and is oriented upwards (in the z-direction). This upward orientation is crucial as it defines the intended direction in which the surface will interact with the vector field \( \mathbf{F}\). - The dot product \( \mathbf{F} \cdot \mathbf{n} \) decides how much the vector field affects the surface. A higher result suggests a stronger effect or more alignment (here, it evaluates to 3, indicating moderate alignment in the positive z-direction).- The normal vector helps in projecting vector fields onto a surface, enabling engineers and scientists to determine how forces like wind or electromagnetism might act upon material structures.Thus, understanding normal vectors is key to accurately performing operations like computing surface integrals, as it guides the orientation and thereby the overall interaction between vectors and surfaces.