Problem 371 - Calculus Volume 3

Question

Circle \(C\) in plane \(x+y+z=8\) has radius 4 and center \((2,3,3) .\) Evaluate \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} \quad\) for \(F=\langle 0,-z, 2 y\rangle, \quad\) where \(C\) has a counterclockwise orientation when viewed from above.

Step-by-Step Solution

Answer

The line integral is \(-\frac{16 \pi}{\sqrt{3}}\).

Step 1: Identify the Condition for Green's Theorem

First, note that Green's Theorem applies only in the plane. Our scenario involves a circle in the plane given by \(x + y + z = 8\), which is not the same as the standard xy-plane. Therefore, we need to consider the surface that \(C\) bounds to use Stokes' Theorem instead of Green's Theorem.

Step 2: Modify the Field for Stokes' Theorem

Stokes' Theorem relates the surface integral of the curl of a vector field to a line integral over its boundary. We need to find the curl of \(\mathbf{F} = \langle 0, -z, 2y \rangle\).

Step 3: Compute the Curl of F

The curl of a vector field \( \mathbf{F} = \langle P, Q, R \rangle \) is given by \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \). For \( \mathbf{F} = \langle 0, -z, 2y \rangle \), this becomes \( abla \times \mathbf{F} = \left(0 - 0, 0 - 0, -1 - 0\right) = \langle 0, 0, -1 \rangle \).

Step 4: Evaluate the Surface Integral

Now, using Stokes' Theorem \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} (abla \times \mathbf{F}) \cdot d\mathbf{S} \), where \(d\mathbf{S}\) is the surface element of the plane \(x + y + z = 8\). The unit normal vector of this plane can be approximated to be \( \mathbf{n} = \langle 1, 1, 1 \rangle / \sqrt{3} \). The intersection of the plane with the circle can be parametrize as needed, but crucially the constant \(\left(abla \times \mathbf{F}\right) \cdot \mathbf{n}\) simplifies to evaluating \(\langle 0, 0, -1 \rangle \cdot \langle 1, 1, 1 \rangle / \sqrt{3} = -1 / \sqrt{3} \).

Step 5: Calculate the Area of the Surface

The surface area that \(\mathbf{n}\) spans in relation to \(C\) is part of a unit sphere intersected by our plane. To find the area, compensate using the geometry of the sphere. The radius of \(C\) is \(4\), so the circle's area is \( \pi \times 4^2 = 16\pi \).

Step 6: Final Calculation Using Constants Derived

By Stokes' Theorem, substitute the values to get the integral: \(\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{S} abla \times \mathbf{F} \cdot d\mathbf{S} = \left(-\frac{1}{\sqrt{3}} \right) \cdot (16\pi) = -\frac{16 \pi}{\sqrt{3}} \).

Key Concepts

Line Integral

A line integral helps us to calculate a kind of cumulative effect along a curve. Imagine you have a ski slope described by a function, and you're curious about its total steepness over the whole path. That's when a line integral comes into play! It's a way to add up values of a vector field along a specific path.

The beauty of line integrals is their flexibility. Depending on what you need, they handle functions giving you either vectors or scalar values:

• Vector-valued functions: This is typical in physics, for example, calculating work done by a force.
• Scalar functions: Useful for length, mass, or entropy over a path, applicable in various scientific contexts.

To compute line integrals, you'll often see them expressed as an integral over a path, like \(\oint_{C} \mathbf{F} \cdot d \mathbf{r} \), where \( \mathbf{F} \cdot d\mathbf{r} \) describes how the field aligns with the movement along \(C\).

Surface Integral

A surface integral extends the concept of a line integral to two dimensions. Instead of a path, we're looking at a surface. Imagine you want to know the total amount of water hitting a tilted roof during a rainstorm. By using surface integrals, you can sum up the effects across that entire area.

In the mathematical arena:

• Surface integrals for scalar fields: They compute values like the total charge or strange fluid qualities across a surface.
• Surface integrals for vector fields: These are key for understanding flux — measuring how much of a substance passes through a given surface.

The calculated flow across a surface is expressed as \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} \), where \( d\mathbf{S} \) is the small piece of the surface area, and \( \mathbf{F} \cdot d\mathbf{S} \) captures how the vector field \( \mathbf{F} \) interacts with this tiny surface portion.

Vector Field

A vector field is like a field of arrows spread over an area that tells you exactly where and how strongly something acts at each point. Think of a map where every point has a tiny arrow showing wind direction and speed — that's a classic vector field.

This concept is highly applicable:

• Fluids: For instance, the flow direction and strength of water or air at various points.
• Forces: Gravitational, electric, or magnetic fields that exert influences on objects.

The vector field in this exercise, \( \mathbf{F} = \langle 0, -z, 2y \rangle \), represents values dependent on position components \(y\) and \(z\). This indicates how \(\mathbf{F}\) behaves when interacting with objects and paths within the space.

Curl of a Vector Field

The curl of a vector field measures how much the field wants to rotate or swirl around a point. It's the spinning strength of a field at each location, like observing a whirlpool in a pond.

In formulas, the curl is denoted as \( abla \times \mathbf{F} \). Here's the practical side:

• Vorticity in fluids: Used to describe the local spinning motion of the fluid.
• Rotational forces: Essential in electromagnetism and fluid dynamics.

In our scenario, we compute the curl for \( \mathbf{F} = \langle 0, -z, 2y \rangle \). Hence, \( abla \times \mathbf{F} = \langle 0, 0, -1 \rangle \) becomes the tool to explore the rotational aspect of this field as we apply Stokes' Theorem.