Problem 4 - 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 \(C : \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi\) $$\mathbf{F}=-x^{2} y \mathbf{i}+x y^{2} \mathbf{j}$$

Step-by-Step Solution

Answer

Both integrals evaluate to \( \frac{\pi a^4}{2} \), confirming Green's Theorem.

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 \( C \). This is expressed as: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] where \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \).

Step 2: Identify M and N

For the given vector field \( \mathbf{F} = -x^2 y \mathbf{i} + x y^2 \mathbf{j} \), identify \( M = -x^2 y \) and \( N = x y^2 \).

Step 3: Compute \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \)

Calculate the partial derivatives: \( \frac{\partial N}{\partial x} = y^2 \) and \( \frac{\partial M}{\partial y} = -x^2 \).

Step 4: Express the Double Integral

Substitute into Green's Theorem: \[ \iint_{R} \left( y^2 - (-x^2) \right) \, dA = \iint_{R} (x^2 + y^2) \, dA \]. Since \( R \) is a disk of radius \( a \), this simplifies to \[ \iint_{R} r^2 \, dr \, d\theta \] in polar coordinates.

Step 5: Evaluate the Double Integral

In polar coordinates, \( r^2 = x^2 + y^2 \) and \( dA = r \, dr \, d\theta \). The limits for \( r \) are from 0 to \( a \) and for \( \theta \) from 0 to \( 2\pi \). Thus, \[ \iint_{R} r^2 \, dA = \int_{0}^{2\pi} \int_{0}^{a} r^3 \, dr \, d\theta \]. Evaluate the inner integral: \[ \int_{0}^{a} r^3 \, dr = \frac{a^4}{4} \]. Now, evaluate the outer integral: \[ \int_{0}^{2\pi} \frac{a^4}{4} \, d\theta = \frac{a^4}{4} \cdot 2\pi = \frac{\pi a^4}{2} \].

Step 6: Evaluate the Line Integral

The line integral around the circle \( C \) is: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \oint_{0}^{2\pi} (-x^2 y \frac{dx}{dt} + x y^2 \frac{dy}{dt}) \, dt \]. Parametrize \( C \) using \( x = a \cos t \) and \( y = a \sin t \) with \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = a \cos t \). Substitute these into the integral and simplify: \[ \oint_{0}^{2\pi} (-a^2 \cos^2 t \cdot a \sin t \cdot (-a \sin t) + a^2 \cos t \cdot a^2 \sin^3 t) \, dt \]. Simplify further to \[ \oint_{0}^{2\pi} a^4 \cos t \sin^3 t \, dt \].

Step 7: Confirm Equality of Integrals

Both the double integral and the line integral should equal \( \frac{\pi a^4}{2} \) after evaluation. Use symmetry or trigonometric identities to show both integrals compute to the same result as a verification of Green's Theorem.

Key Concepts

Line Integrals and Green's Theorem

Line integrals are a fundamental concept in calculus, used to find the integral of a function along a curve. Imagine if you were walking along a path with a gentle slope. A line integral could help you determine how much effort it takes to walk that path. In mathematics, line integrals help in evaluating functions over curves in a vector field.

In Green's Theorem, the line integral is considered around a closed curve on a plane. This theorem provides a fascinating link between the line integral around a curve and the double integral over the region bounded by the curve. The equation for Green's Theorem can be written as: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]
Here:

• \( C \) is the closed curve.
• \( \mathbf{F} \) is a vector field, which can be written as \( M \mathbf{i} + N \mathbf{j} \).
• \( d\mathbf{r} \) represents a small segment of the curve.

Green's Theorem essentially says, "The total shortfall around the path equals the sum of the whirlpools inside." This connection helps in solving complex line integrals using simpler double integrals, provided the conditions of the system are met.

Understanding Double Integrals

Double integrals allow you to add up values over a two-dimensional region, making them very useful in physics and engineering. They work by slicing a bean-shaped area into tiny pieces and adding up a value for each piece. Imagine sprinkling a field with seeds evenly and using double integrals to compute the total number of seeds over that field. This is analogous to how double integrals aggregate data over an area.

In the context of Green's Theorem, we consider a double integral over a disk-shaped region \( R \), specifically evaluating the expression \( (x^2 + y^2) \) when it’s transformed into polar coordinates. The switch to polar coordinates is done because they are simpler to work with for circular domains:\[ \iint_{R} (x^2 + y^2) \, dA = \iint_{R} r^2 \, dr \, d\theta \]

• \( r \) is the radius from the center of the circle.
• \( \theta \) goes from 0 to \( 2\pi \), covering the whole circle.

To evaluate, you perform an inner integral over \( r \) and an outer integral over \( \theta \). This method enables the computation of quantities over complex shapes, akin to painting a target with precise layers of color.

Exploring Vector Fields

Vector fields are like the invisible hands of nature, pushing or pulling along each point in a space. Imagine a field where each grass blade points in a specific direction and length — that’s a vector field. In mathematics, these fields are critical in describing forces, velocities, and other quantities having both magnitude and direction, across various dimensions.

For the vector field \( \mathbf{F} = -x^2 y \mathbf{i} + x y^2 \mathbf{j} \), we break it into its respective \( x \) and \( y \) components:

• \( M = -x^2 y \)
• \( N = x y^2 \)

These components tell us how strong and which way the force moves at every point in our region. Analyzing the partial derivatives of these components is crucial in applying Green's Theorem. By doing it, we can understand the curl or tendency to rotate within the field, which helps simplify complex spatial computations.
Vector fields are not just imaginary concepts; they underpin many phenomena in real life, from electrical circuits to fluid dynamics, offering insights into interactions and behaviors at multiple levels.