Problem 2 - Thomas Calculus

Question

In Exercises \(1-4,\) verify the conclusion of Green's Theorem by evaluating both sides of Equations \((3)\) and \((4)\) for the field \(\mathbf{F}=M \mathbf{i}+N \mathbf{j}\) . Take the domains of integration in each case to be the disk \(R : x^{2}+y^{2} \leq a^{2}\) and its bounding circle \(C : \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi\) $$\mathbf{F}=y \mathbf{i}$$

Step-by-Step Solution

Answer

Both integrals equal \(-\pi a^2\), confirming Green's Theorem.

Step 1: Define the Field Components

The vector field is given as \(\mathbf{F} = y \mathbf{i} + 0 \mathbf{j}\). Therefore, \(M = y\) and \(N = 0\).

Step 2: Calculate the Curl in the Plane

Green's Theorem (equation \((3)\)) relates the line integral around a simple, closed curve \(C\) to a double integral over the plane region \(R\) it encloses: \(\oint_{C} (M\, dx + N\, dy) = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA\). Calculate the partial derivatives: \(\frac{\partial N}{\partial x} = 0\) and \(\frac{\partial M}{\partial y} = \frac{\partial y}{\partial y} = 1\). Thus, \(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = -1\).

Step 3: Evaluate the Double Integral

The double integral over the region \(R\) is \(\iint_{R} (-1) \, dA = -\iint_{R} dA\). The region \(R\) is a disk with radius \(a\), and its area is \(\pi a^2\). Thus, \(-\iint_{R} dA = -\pi a^2\).

Step 4: Evaluate the Line Integral

The line integral is given by \(\oint_{C} M \, dx + N \, dy = \oint_{C} y \, dx + 0 \, dy\). Parameterize the curve \(C\) with \(x = a \cos t\) and \(y = a \sin t\), with \(dx = -a \sin t \, dt\). Now, substitute these into the line integral: \[ \oint_{C} a \sin t (-a \sin t) \, dt = -a^2 \oint_{0}^{2\pi} \sin^2 t \, dt.\]

Step 5: Integrate the Sine Squared Term

Using the identity \(\sin^2 t = \frac{1}{2} (1 - \cos 2t)\), we integrate:\[ \oint_{0}^{2\pi} \frac{1}{2} (1 - \cos 2t) \, dt = \frac{1}{2} \left[ t - \frac{1}{2} \sin 2t \right]_{0}^{2\pi} = \pi.\]Thus, the line integral is \(-a^2 \pi\).

Step 6: Conclude Verification by Comparing Results

From Step 3, the double integral gave \(-\pi a^2\). From Step 5, the line integral gave \(-\pi a^2\). Since both results are equal, the conclusion of Green's Theorem is verified.

Key Concepts

Vector Field

In the context of Green's Theorem, a vector field is a crucial concept. A vector field assigns a vector to every point in a plane or space, which can be represented in two dimensions as \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \). Here, \( M \) and \( N \) are functions of coordinates \( x \) and \( y \).
For example, in our exercise, the vector field is given by \( \mathbf{F} = y \mathbf{i} + 0 \mathbf{j} \). This means the vectors only have a component in the \( \mathbf{i} \) direction, and their length varies linearly with \( y \). There is no component in the \( \mathbf{j} \) direction.
- **Components**: The field is described by two functions, \( M = y \) and \( N = 0 \).- **Direction**: The vector's direction is horizontal, pointing along the \( x \)-axis.Understanding vector fields is fundamental when applying Green's Theorem because they help define what curves \( C \) and regions \( R \) are interacting with in the plane.

Line Integral

A line integral is a critical tool in understanding how a vector field interacts with a curve. It's like tracing a curve with a tiny ruler that measures not just distance, but also accounts for forces (from vector fields) along the way.
For a given vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), the line integral around a closed curve \( C \) is expressed as:\[ \oint_{C} (M \, dx + N \, dy). \]In our exercise:

• Parameterization: The curve \( C \) is represented in a parametric form with \( x = a \cos t \) and \( y = a \sin t \), where \( 0 \leq t \leq 2 \pi \).
• Substitution: This parameterization is plugged into the expression for the line integral, allowing evaluation over \( t \).
• Integration: The integration of \( a \sin t (-a \sin t) \) from 0 to \( 2\pi \), incorporating the identity \( \sin^2 t = \frac{1}{2}(1 - \cos 2t) \), arrives at \( -a^2 \pi \).

Line integrals allow us to effectively "sum up" the influence of a vector field along a path, which is foundational in verifying Green's Theorem.

Double Integral

While line integrals look at paths, double integrals examine regions. They measure how a vector field affects an entire area. In Green's Theorem, they connect with line integrals, showing the total flux across a region's boundary.
The double integral is written over a region \( R \) in the plane as\[ \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA. \]In our situation:

• The derivative \( \frac{\partial N}{\partial x} = 0 \) and \( \frac{\partial M}{\partial y} = 1 \), leading to the expression \( -1 \), simplifying our double integral to \( -\iint_{R} \, dA \).
• The region \( R \) is a disk with radius \( a \), meaning the area is \( \pi a^2 \).
• Calculating \( -\iint_{R} \, dA \), we find \( -\pi a^2 \), showing how the entire field influences the region \( R \).

By connecting area effects (double integrals) and boundary effects (line integrals), Green's Theorem validates their equivalence, which is affirmed as both evaluate to \( -\pi a^2 \).