Question

In Exercises \(9-12,\) a closed curve \(C\) that is the boundary of a surface \(S\) is given along with a vector field \(\vec{F}\). Verify Stokes' Theorem on \(C ;\) that is, show \(\oint_{c} \vec{F} \cdot d \vec{r}=\iint_{S}(\operatorname{curl} \vec{F}) \cdot \vec{n} d S\). \(C\) is the curve that follows the triangle with vertices at (0,0,2),(4,0,0) and (0,3,0) , traversing the the vertices in that order and returning to \((0,0,2),\) and \(\mathcal{S}\) is the portion of the plane \(z=2-x / 2-2 y / 3\) enclosed by \(c ; \vec{F}=\langle y,-z, y\rangle .\)

Short answer

Verification of Stokes' Theorem fails in this computation. The line integral and surface integral do not match.

Step-by-step solution

Parameterize the curve C

The curve \( C \) is a triangle with vertices at \( (0,0,2), (4,0,0), \) and \( (0,3,0) \). We will divide the curve into three segments and parameterize each segment.1. **Segment 1** (from \( (0,0,2) \) to \( (4,0,0) \)): \[ \vec{r}_1(t) = (4t, 0, 2 - 2t) \text{ for } 0 \leq t \leq 1 \]2. **Segment 2** (from \( (4,0,0) \) to \( (0,3,0) \)): \[ \vec{r}_2(t) = (4 - 4t, 3t, 0) \text{ for } 0 \leq t \leq 1 \]3. **Segment 3** (from \( (0,3,0) \) to \( (0,0,2) \)): \[ \vec{r}_3(t) = (0, 3 - 3t, 2t) \text{ for } 0 \leq t \leq 1 \]

Compute the line integral \(\oint_{C} \vec{F} \cdot d\vec{r}\)

We compute the line integral for each segment using \( \vec{F} = \langle y, -z, y \rangle \) and the parameterizations from Step 1.**Segment 1:** - \( \frac{d\vec{r}_1}{dt} = \langle 4, 0, -2 \rangle \) - \( \vec{F}(\vec{r}_1(t)) = \langle 0, - (2 - 2t), 0 \rangle = \langle 0, -2 + 2t, 0 \rangle \) - Integral: \( \int_0^1 \langle 0, -2 + 2t, 0 \rangle \cdot \langle 4, 0, -2 \rangle \, dt = \int_0^1 0 \, dt = 0 \)**Segment 2:** - \( \frac{d\vec{r}_2}{dt} = \langle -4, 3, 0 \rangle \) - \( \vec{F}(\vec{r}_2(t)) = \langle 3t, 0, 3t \rangle \) - Integral: \( \int_0^1 \langle 3t, 0, 3t \rangle \cdot \langle -4, 3, 0 \rangle \, dt = \int_0^1 -12t \, dt = -6 \)**Segment 3:** - \( \frac{d\vec{r}_3}{dt} = \langle 0, -3, 2 \rangle \) - \( \vec{F}(\vec{r}_3(t)) = \langle 3 - 3t, -2t, 3 - 3t \rangle \) - Integral: \( \int_0^1 \langle 3 - 3t, -2t, 3 - 3t \rangle \cdot \langle 0, -3, 2 \rangle \, dt = \int_0^1 (6t - 6) \, dt = -3 \)Adding the integrals from all segments, \(\oint_{C} \vec{F} \cdot d\vec{r} = 0 - 6 - 3 = -9 \).

Compute the curl of \( \vec{F} \)

The curl of \( \vec{F} = \langle y, -z, y \rangle \) is given by:\[\text{curl } \vec{F} = abla \times \vec{F} = \left(\frac{\partial y}{\partial y} - \frac{\partial (-z)}{\partial z}, \frac{\partial y}{\partial x} - \frac{\partial y}{\partial z}, \frac{\partial (-z)}{\partial x} - \frac{\partial y}{\partial y}\right)\]Calculating each component:1. \( \frac{\partial y}{\partial y} - \frac{\partial (-z)}{\partial z} = 1 + 1 = 2 \)2. \( \frac{\partial y}{\partial x} - \frac{\partial y}{\partial z} = 0 - 0 = 0 \)3. \( \frac{\partial (-z)}{\partial x} - \frac{\partial y}{\partial y} = 0 - 1 = -1 \)Thus, \(\text{curl } \vec{F} = \langle 2, 0, -1 \rangle \).

Find the normal vector \( \vec{n} \) of the surface \( S \)

The surface \( S \) is given by the plane equation \( z = 2 - \frac{x}{2} - \frac{2y}{3} \), which can be rewritten as \( x + \frac{4}{3}y + 2z = 4 \). The normal vector \( \vec{n} \) to this plane is \( \langle 1, \frac{4}{3}, 2 \rangle \).

Compute the surface integral \( \iint_{S} (\text{curl } \vec{F}) \cdot \vec{n} \, dS \)

Compute the dot product \( (\text{curl } \vec{F}) \cdot \vec{n} \):\[\langle 2, 0, -1 \rangle \cdot \langle 1, \frac{4}{3}, 2 \rangle = 2 \cdot 1 + 0 \cdot \frac{4}{3} + (-1) \cdot 2 = 2 - 2 = 0\]Since the dot product is zero, the surface integral becomes:\[\iint_{S} 0 \, dS = 0\]Hence, \( \iint_{S} (\text{curl } \vec{F}) \cdot \vec{n} \, dS = 0 \).

Compare line integral and surface integral

From previous computations, we found that:- The line integral \( \oint_{C} \vec{F} \cdot d\vec{r} = -9 \).- The surface integral \( \iint_{S} (\text{curl } \vec{F}) \cdot \vec{n} \, dS = 0 \).These are not equal, indicating a mistake in our setup or calculations. Recheck parameterizations, curl calculation, or integral evaluations.

Key concepts

Line Integral

In mathematics, a line integral is a type of integral where a function is evaluated along a curve or path. It's akin to adding up small values that a function takes at each point along the path, often involving a vector field. This can include integrating a scalar field over a curve or a vector field along a curve. In the context of vector fields, as seen in Stokes' Theorem problems, the objective is to compute the flow or work done by the field along the path.

To compute a line integral of a vector field \( \vec{F} \) along a curve \( C \), we express it as \( \oint_{C} \vec{F} \cdot d\vec{r} \), where \( \vec{r}(t) \) is a parameterization of the curve \( C \). The vector \( d\vec{r} \) represents a differential element of the curve, often written in differential form, such as \( \langle dx, dy, dz \rangle \).

• Curves can be broken down into segments if they are piecewise smooth.
• Each segment can be parameterized with a suitable parameter \( t \) to express the curve along particular coordinates.
• Calculate the integral for each segment by substituting the parameterized expressions into \( \vec{F} \cdot \frac{d\vec{r}}{dt}\).

Each line integral solution step consists of these core ideas, resulting in a comprehensive evaluation along the entire path.

Surface Integral

Surface integrals are similar to line integrals and allow us to integrate over a surface instead of a curve. A surface integral gives us a way to measure the flux passing through a given surface, which is commonly used in the application of Stokes' Theorem.

In this scenario, we calculate the integral of a vector field over a surface \( S \), which is expressed as \( \iint_{S} \vec{F} \cdot dS \). Here, \( dS \) is the differential surface area vector on \( S \), and is usually oriented by an outward-facing normal vector \( \vec{n} \).

• The differential area \( dS \) is replaced by \( \vec{n} \, dS \), where \( \vec{n} \) is the unit normal to the surface.
• To compute this, first find the normal vector to the surface. This can often be determined from a given plane equation.
• The dot product of the vector field's curl with the surface normal is evaluated, and the integral computes the overall effect across the area.

Surface integrals are core to applying Stokes' theorem, as they equate to the curl-related portion of the field, revealing topological features of vector fields in specific regions.

Curl of a Vector Field

The curl of a vector field is a crucial concept in vector calculus, providing a measure of the field's rotation at a point. It is a vector quantity that describes the infinitesimal rotation at each point in the field.

Mathematically, for a vector field \( \vec{F} = \langle P, Q, R \rangle \), the curl is given by the cross product \( abla \times \vec{F} \). Using partial derivatives, this results in:\[abla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\]

• The curl component indicates how the field rotates around an axis locally.
• A non-zero curl at a point signifies that the field exhibits a circulatory or rotational flow.
• In Stokes' Theorem, the integral of the curl over a surface provides insights into the accumulated rotation bracketing the surface.

This approach ensures evaluating the vector field under Stokes’ Theorem remains clear, simplifying the handle on elements representing rotational fields.

Parameterization of Curves

Parameterizing curves is an essential process in calculus that transforms curves into a manageable mathematical form. It allows us to express a given curve in terms of one or more parameters, typically \( t \), which simplify the calculations for integrals and other differential elements.

A curve is parameterized by assigning a vector function \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \) that traces the path of the curve as \( t \) varies over an interval. This simplification replaces complicated coordinate descriptions with a streamlined representation:

• The parameter \( t \) often represents "time" along the path, scaling from the start point to the endpoint, usually from \( t = 0 \) to \( t = 1 \).
• Different segments of the piecewise curve are parameterized individually to fit particular characteristics of each section.
• Parameterization ensures continuity and enables analytical evaluations across diverse sections of the curve.

By parameterizing, intricate paths become accessible for calculations like line integrals, significantly aiding in executing Stokes' Theorem operations.