Problem 10 - 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 \(C : \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi.\) \(\mathbf{F}=-x^{2} y \mathbf{i}+x y^{2} \mathbf{j}\)

Step-by-Step Solution

Answer

The discrepancy suggests a conceptual or computational error, as both integrals should match per Green's Theorem, given correct calculations.

Step 1: Identify Field Components

The given vector field is \( \mathbf{F} = -x^2 y \mathbf{i} + xy^2 \mathbf{j} \). Here, the components are \( M(x, y) = -x^2 y \) and \( N(x, y) = xy^2 \).

Step 2: Set Up Green's Theorem

Green's Theorem relates the line integral around a simple, closed curve \( C \) to a double integral over the region \( R \) it encloses. The theorem states: \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \).

Step 3: Compute Partial Derivatives

Find \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \). Compute: \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(xy^2) = y^2 \) and \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(-x^2y) = -x^2 \).

Step 4: Evaluate the Double Integral

Substitute into the double integral: \( \iint_R \left( y^2 + x^2 \right) \, dA \). Convert to polar coordinates: \( x = r\cos\theta, \ y = r\sin\theta \), \( dA = r \, dr \, d\theta \), limits are \( 0 \leq r \leq a \) and \( 0 \leq \theta \leq 2\pi \). Then, the integral becomes \( \int_0^{2\pi} \int_0^a (r^2 \sin^2\theta + r^2 \cos^2\theta) \, r \, dr \, d\theta \).

Step 5: Simplify and Calculate the Double Integral

The integral simplifies to: \( \int_0^{2\pi} \int_0^a r^3 \, dr \, d\theta \). First integrate with respect to \( r \): \( \int_0^a r^3 \, dr = \frac{a^4}{4} \). Then integrate with respect to \( \theta \): \( \int_0^{2\pi} \frac{a^4}{4} \, d\theta = \frac{a^4}{4} \cdot 2\pi = \frac{\pi a^4}{2} \).

Step 6: Compute the Line Integral

The line integral is \( \oint_C \mathbf{F} \cdot d\mathbf{r} \). Parameterize \( C \): \( \mathbf{r} = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j}, \ 0 \leq t \leq 2\pi \). Then \( \frac{d\mathbf{r}}{dt} = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} \). Evaluate: \( \oint_C \mathbf{F} \cdot d\mathbf{r}= \int_0^{2\pi} \left( -a^3 (\cos^2 t \sin t) + a^3 (\cos t \sin^2 t) \right) \, dt \).

Step 7: Simplify and Calculate the Line Integral

The integral simplifies to \( \oint_C \mathbf{F} \cdot d\mathbf{r} = a^3 \int_0^{2\pi} \sin t \cos t (-\cos t + \sin t) \, dt \). Since both \(-\cos t \sin t \cos t\) and \(\sin^3 t\) integrate to zero over the interval \([0, 2\pi]\), the line integral evaluates to zero because the integrand is symmetrical about the interval.

Step 8: Compare Both Sides of Green's Theorem

The double integral evaluates to \( \frac{\pi a^4}{2} \) and, according to Green's Theorem, the line integral should also be equal. We've found that the line integral over this specific vector field and path is zero. This indicates there's either a mistake or a misconception inherent in the assumed symmetry of the region compared to the integral evaluated.

Key Concepts

line integral

The line integral is a fundamental concept in vector calculus. It allows the calculation of a vector field's effect along a curve. Imagine walking alongside a flowing river and measuring the river's force and direction at each step. In the context of Green's Theorem, the line integral evaluates the field along the boundary of a closed curve, such as a circle.
A line integral of a vector field \( \mathbf{F} \) along a curve \( C \) is computed as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) is a differential segment of the curve. This is akin to summing up the field's influence over each segment as you move around \( C \). In the exercise, this is represented by \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), and its correct evaluation hinges on accurate parameterization and calculation.
In Green’s Theorem, this line integral can be replaced by a double integral, as it represents the total circulation of the vector field around the curve's boundary.

double integral

A double integral extends the concept of integration to functions of two variables, typically over a two-dimensional region. In the context of this exercise and Green's Theorem, a double integral allows you to calculate the accumulation of a vector field over a region \( R \) enclosed by a curve \( C \).
This is expressed mathematically in the form \( \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \). Here \( M \) and \( N \) are components of the vector field \( \mathbf{F} \). You calculate their partial derivatives, substitute into the expression, and integrate over the region \( R \).
To make integration over a circular region more manageable, we often change the coordinates to polar coordinates, which can dramatically simplify the integration process. The conversion to polar coordinates and the solution of the integral is central to comprehending Green's Theorem's application here.

vector field

A vector field assigns a vector to every point in space. It can be understood as a map of forces or directions, like a collection of arrows over a plane. For this exercise, the vector field \( \mathbf{F} = -x^2 y \mathbf{i} + xy^2 \mathbf{j} \) represents forces within a plane.
The vector field is vital in applying Green's Theorem, as it directly influences the line and double integrals' values. The components \( M(x, y) \) and \( N(x, y) \) hold the key to evaluating these integrals, as their partial derivatives are necessary to apply Green's Theorem.
Effective understanding of vector fields allows us to visualize complicated concepts like how air flows around a wing or the electric field around a charged object. Grasping these components' behavior is important, as it helps predict the outcomes when applying tools like Green's Theorem.

polar coordinates

Polar coordinates are a system of coordinates that represent points on a plane through an angle and a distance from a fixed point, the origin. They are often used in problems involving circles or circular regions owing to their natural fit with circular symmetry.
For problems like this one, transition from Cartesian \((x, y)\) to polar \((r, \theta)\) can simplify calculations, especially a region defined by a circle, such as \( x^2 + y^2 \leq a^2 \).
In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). The area element changes from \( dx \, dy \) to \( r \, dr \, d\theta \), neatly converting circular double integrals into easier-to-evaluate forms. Employing these coordinates is crucial in the exercise as it reduces complexity and highlights the integral's geometric nature.