Problem 3 - 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 \(C : \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi\) $$\mathbf{F}=2 x \mathbf{i}-3 y \mathbf{j}$$

Step-by-Step Solution

Answer

Both integrals are zero, confirming Green's Theorem's conclusion.

Step 1: Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \). It states that \( \oint_C (M \, dx + N \, dy) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \).

Step 2: Identify the Field Components

The given vector field is \( \mathbf{F} = 2x\mathbf{i} - 3y\mathbf{j} \), where \( M = 2x \) and \( N = -3y \).

Step 3: Compute Partial Derivatives

Find \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \). Here, \( \frac{\partial N}{\partial x} = 0 \) because \( N = -3y \) doesn't depend on \( x \), and \( \frac{\partial M}{\partial y} = 0 \) because \( M = 2x \) doesn't depend on \( y \).

Step 4: Evaluate the Double Integral

Substitute \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \) into the double integral: \( \iint_R (0 - 0) \, dA = 0 \). Since both partial derivatives are zero, the double integral over the region \( R \) also equals zero.

Step 5: Parametrize the Curve \( C \)

The curve \( C \) is the circle given by \( \mathbf{r}(t) = a \cos t \mathbf{i} + a \sin t \mathbf{j} \), for \( 0 \leq t \leq 2\pi \). The derivatives \( dx \) and \( dy \) are \( -a \sin t \, dt \) and \( a \cos t \, dt \) respectively.

Step 6: Evaluate the Line Integral

The line integral is \( \oint_C (2x\,dx + (-3y)\,dy) \). Substitute \( x = a\cos t \) and \( y = a\sin t \) into the integral: \( \int_0^{2\pi} [2(a\cos t)(-a\sin t) - 3(a\sin t)(a\cos t)] \, dt \). This expression simplifies to zero since the two terms cancel each other out.

Step 7: Compare Both Results

Both the double integral and the line integral around \( C \) are zero. This confirms that the conclusion of Green's Theorem holds: \( \oint_C (M \, dx + N \, dy) = 0 \) and \( \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA = 0 \).

Key Concepts

Line Integral

The concept of a line integral is essential in vector calculus, where it represents the integral of a function along a curve. Imagine a vector field as a flow of water, and you are interested in the amount of flow along a certain path. The line integral helps quantify this flow.
The line integral of a vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \) along a closed curve \( C \) is expressed as:

• \( \oint_C (M \, dx + N \, dy) \)

The integral sums up the contributions of the field along each point of the curve. In practice, we often employ parameterizations. For example, the curve \( C \) could be parameterized by a vector function \( \mathbf{r}(t) \). Substituting into the integral, we calculate the total effect of the field along the path.
This technique is crucial when dealing with fields in areas like electromagnetism or fluid dynamics. When applying Green's Theorem, the line integral is computed and compared with the double integral over a region, as they offer different viewpoints into analyzing the behavior of vector fields.

Double Integral over a Region

A double integral extends the concept of integration from one dimension to two dimensions. It allows us to calculate things like volume under a surface or mass of a lamina (a thin plate) covering a region. When referring to Green's Theorem, the double integral relates to the circulation or curl of a vector field over a certain area.
The expression:

• \( \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \)

sums up values over the region \( R \), which is bounded by the curve \( C \). The terms \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \) are the partial derivatives of the components of \( \mathbf{F} \). These represent the rotational effect of the field, or how the field swirls within the region. If both partial derivatives are zero, as in the provided exercise, it indicates no net rotation,"" implying that the field doesn't swirl within the circle.
Applying a double integral makes it possible to evaluate many physical quantities over surfaces, such as center of mass and pressure, and is integral in verifying Green's Theorem.

Partial Derivatives

Partial derivatives enable us to understand how a function changes as one of its variables changes while keeping other variables constant. They are foundational in multivariable calculus, essential for Green's Theorem, as they measure the rate of change of a variable field based on spatial coordinates.
Given a vector field \( \mathbf{F} = 2x \mathbf{i} - 3y \mathbf{j} \), the partial derivative \( \frac{\partial N}{\partial x} \) regards the change in the \( y \)-component as \( x \) varies, and \( \frac{\partial M}{\partial y} \) does the reverse. In this exercise:

• \( \frac{\partial N}{\partial x} = 0 \)
• \( \frac{\partial M}{\partial y} = 0 \)

These zero values indicate that neither component is sensitive to changes in the other variable, thereby simplifying the double integral over region \( R \) to zero.
This property highlights that the field components act independently with respect to each other's primary variables and showcases the interplay of differential calculus in evaluating vector fields.