Problem 7 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Both integrals evaluate to \(2\pi a^2\), verifying Green's Theorem.

Step 1: Define Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) and a double integral over the plane region \(R\) bounded by \(C\). It is given by\[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA, \] where \(\mathbf{F} = M \mathbf{i} + N \mathbf{j}\).

Step 2: Identify M and N

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

Step 3: Calculate Partial Derivatives

Compute the partial derivatives needed for Green's Theorem:1. \(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x) = 1\)2. \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(-y) = -1\).

Step 4: Evaluate Double Integral

Substitute the partial derivatives into the double integral:\[ \iint_R (1 - (-1)) \, dA = \iint_R 2 \, dA. \]The area of the disk \(R\) is \(\pi a^2\), thus the integral becomes \(2 \times \pi a^2\).

Step 5: Parameterize and Evaluate Line Integral

Parametrize the circle \(C\) using \( \mathbf{r}(t) = a \cos t \, \mathbf{i} + a \sin t \, \mathbf{j} \). Then, \( d\mathbf{r} = (-a \sin t \, \mathbf{i} + a \cos t \, \mathbf{j}) \, dt \).Evaluate the line integral: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} [-a \sin t, a \cos t] \cdot [-a \sin t, a \cos t] \, dt = \int_0^{2\pi} (a^2 \sin^2 t + a^2 \cos^2 t) \, dt. \]This simplifies to \(a^2 \int_0^{2\pi} (\sin^2 t + \cos^2 t) \, dt = a^2 \int_0^{2\pi} 1 \, dt = a^2 [t]_0^{2\pi} = 2 \pi a^2\).

Step 6: Verify the Equivalence

Both the line integral around \(C\) and the double integral over \(R\) evaluate to \(2 \pi a^2\). This confirms the conclusion of Green's Theorem.

Key Concepts

Line Integrals

Line integrals are a fundamental concept in vector calculus and are used to evaluate functions along a curve. To better understand, imagine you are walking along a path and measuring certain quantities, like temperature or wind speed. The sum of all these measurements gives you a line integral.

In the context of Green's Theorem, a line integral calculates the circulation of a vector field along a closed curve. Mathematically, it's expressed as:

• The integral over a curve \(C\) of a vector field \(\mathbf{F} = M \mathbf{i} + N \mathbf{j}\), denoted as \(\oint_C \mathbf{F} \cdot d\mathbf{r}\).
• To evaluate this, you need a parameterization of the curve and a vector dot product along each small piece \(d\mathbf{r}\) of the curve.
• In this example, the curve \(C\) is a circle, parameterized using trigonometric functions of \(t\), which allows us to compute the line integral from \(0\) to \(2\pi\).

Using these steps, the line integral becomes a powerful tool for discovering information about the field along the path.

Double Integrals

Double integrals extend the concept of integration to functions of two variables, integrating over regions of the plane. Consider them as a method to accumulate quantities over an area, rather than simply along a line or a curve.

In our current example with Green's Theorem:

• The double integral \(\iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA\) combines the region \(R\) bounded by a curve.
• This region is a disk with an area of \(\pi a^2\), and the result of the double integral gives insight into how the vector field behaves over this entire area.
• Calculating double integrals often requires adding contributions from tiny areas, making them challenging but incredibly informative in terms of understanding physical properties across regions.

When used in Green's Theorem, they link the flow around a boundary to the behavior inside, revealing deep connections hidden in the mathematics.

Vector Fields

Vector fields assign a vector to each point in space, often representing things like magnetic fields, fluid flow, or wind speeds. In mathematics, a vector field \(\mathbf{F}=M \mathbf{i}+N \mathbf{j}\) is a function that points in the direction and magnitude at each location.

Here's how it aligns with Green's Theorem:

• The vector field in our exercise, \(\mathbf{F} = -y \mathbf{i} + x \mathbf{j}\), shows how vectors rotate around the origin, creating a circular flow.
• Understanding the components \(M = -y\) and \(N = x\) helps us determine how the field behaves and how forces act at various points.
• These features are crucial for analyzing physical phenomena, allowing us to translate physical observations into mathematical predictions.

Vector fields form the foundation for many theorems, like Green's, which connect the behavior of a field along a curve to its behavior over an area.

Partial Derivatives

Partial derivatives measure how a function changes as its variables change individually. They are essential when dealing with multivariable functions, providing information about a function's slope in any given direction.

When applying Green's Theorem:

• We focus on two specific partial derivatives: \(\frac{\partial N}{\partial x}\) and \(\frac{\partial M}{\partial y}\).
• Calculating \(\frac{\partial N}{\partial x} = 1\) means the component \(N\) increases at a constant rate as \(x\) increases, while \(\frac{\partial M}{\partial y} = -1\) shows \(M\) decreases as \(y\) increases.
• The difference \(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\) connects these changes, spotlighting the rotational or swirling behavior of the vector field.

By examining partial derivatives, we gauge how fields morph throughout a region, crucial for forming connections between local changes and overall phenomena.