Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems \(7-12\), use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=y \mathbf{i}+z \mathbf{j}+x \mathbf{k} ; C\) is the triangular curve with vertices \((0,0,0),(2,0,0)\), and \((0,2,2)\), oriented counterclockwise as viewed from above.

Step-by-Step Solution

Answer

The integral is 0.

Step 1: Understand Stokes's Theorem

Stokes's Theorem relates the line integral of a vector field \( \mathbf{F} \cdot \mathbf{T} \) around a closed curve \( C \) to a surface integral over a surface \( S \) bound by \( C \). It states, \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n}\, dS \), where \( \mathbf{n} \) is the unit normal to the surface \( S \).

Step 2: Calculate the Curl of the Vector Field

Given \( \mathbf{F} = y \mathbf{i} + z \mathbf{j} + x \mathbf{k} \), find the curl, \( abla \times \mathbf{F} \). Use the determinant:\[abla \times \mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \y & z & x\end{vmatrix}\]The result is \( abla \times \mathbf{F} = \mathbf{i} + \mathbf{j} + \mathbf{k} \).

Step 3: Define the Surface and Normal Vector

The curve \( C \) forms a triangular surface. Choose the surface \( S \) as the triangular plane with vertices \((0,0,0)\), \((2,0,0)\), and \((0,2,2)\). Use the cross product to find a normal: vectors \( \mathbf{a} = \langle 2,0,0 \rangle \) and \( \mathbf{b} = \langle 0,2,2 \rangle \) yield \( \mathbf{a} \times \mathbf{b} = \langle 0, -4, 4 \rangle \) which simplifies to \( \langle 0, -1, 1 \rangle \). The unit normal is \( \mathbf{n} = \frac{1}{\sqrt{2}}\langle 0,-1,1 \rangle \).

Step 4: Set Up the Surface Integral

The surface integral is \( \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS \). Substitute \( abla \times \mathbf{F} = \langle 1,1,1 \rangle \) and \( \mathbf{n} = \frac{1}{\sqrt{2}}\langle 0,-1,1 \rangle \). \((abla \times \mathbf{F}) \cdot \mathbf{n} = 1 \times 0 + 1 \times (-1) + 1 \times 1 = 0\).

Step 5: Evaluate the Surface Integral

Since the dot product is zero, the surface integral \( \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS = 0 \). Therefore, \( \oint_{C} \mathbf{F} \cdot \mathbf{T} d s = 0 \), according to Stokes's Theorem.

Key Concepts

Vector Field

A vector field is a function that assigns a vector to each point in space. In this exercise, the vector field is given by \( \mathbf{F} = y \mathbf{i} + z \mathbf{j} + x \mathbf{k} \). This means:

• At each point \((x, y, z)\) in space, the vector \( \langle y, z, x \rangle \) is assigned.
• The components of the vector field are functions of spatial coordinates.

To understand vector fields, visualize a vector attached to each point in space. These vectors can vary in magnitude and direction depending on the location in space.
In calculus, vector fields are used to model force fields, fluid flow, and more. Understanding how vectors change through space is crucial for vector calculus applications.

Curl Calculation

The curl of a vector field provides a measure of the field's rotation at a point. For the vector field \( \mathbf{F} = y \mathbf{i} + z \mathbf{j} + x \mathbf{k} \), we find the curl using:\[\mathbf{abla} \times \mathbf{F} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \y & z & x\end{vmatrix}\]When you calculate this using the determinant, you end up with \( \mathbf{i} + \mathbf{j} + \mathbf{k} \).

• Curl is essentially a vector that describes the rotational tendency or "twist" in the field.
• If the curl is zero, the field is irrotational at that point.

For understanding, imagine a spinning paddlewheel placed in the flow of a vector field. The way it spins is determined by the curl.

Surface Integral

A surface integral allows you to sum up a quantity over a surface in space. For the surface integral in Stokes's Theorem, you compute \( \iint_{S} (\mathbf{abla} \times \mathbf{F}) \cdot \mathbf{n} \, dS \). This involves:

• Finding how much of a vector field rotates across a surface \( S \).
• Integrating this quantity across all points on \( S \).

To apply this, you need to determine a parametrization of the surface \( S \), which in this case is a triangle. Here, \( (\mathbf{abla} \times \mathbf{F}) \cdot \mathbf{n} \) simplifies to zero, suggesting no net rotation across this surface. Surface integrals like this are useful in physics, where they can represent quantities like total flux through a surface.

Line Integral

Line integrals measure the contribution of a vector field along a curve. In the context of Stokes's Theorem, the line integral \( \oint_{C} \mathbf{F} \cdot \mathbf{T} \, ds \) represents the circulation of \( \mathbf{F} \) along a closed path \( C \).

• Line integrals are path-dependent, considering how the vector field interacts with the path's direction.
• This calculation describes how the vector field aligns with and moves along the curve.

For this exercise, using Stokes's Theorem converts the line integral into a surface integral, simplifying the problem based on the geometry of \( C \). Line integrals are essential in fields like electromagnetism, where they can describe work done by a field around a loop.

Unit Normal

A unit normal vector \( \mathbf{n} \) is perpendicular to a surface at a point, and has a length of 1. Finding \( \mathbf{n} \) involves calculating:
1. A normal vector using the cross product of vectors in the plane. 2. Dividing the resultant vector by its magnitude to ensure it is a unit vector.For the triangular surface in this exercise, the vectors \( \mathbf{a} = \langle 2, 0, 0 \rangle \) and \( \mathbf{b} = \langle 0, 2, 2 \rangle \) are cross-multiplied to find \( \mathbf{a} \times \mathbf{b} = \langle 0, -4, 4 \rangle \), which simplifies to \( \langle 0, -1, 1 \rangle \). Then, normalizing gives \( \mathbf{n} = \frac{1}{\sqrt{2}} \langle 0, -1, 1 \rangle \), ensuring the vector is of unit length.

• The unit normal is crucial for computing surface integrals, affecting how much of the field crosses the surface perpendicularly.

Unit normal vectors are fundamental in geometry and physics, particularly in defining orientations and surface properties.