Problem 8 - Thomas Calculus

Question

In Exercises \(5-14,\) use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(F\) and curve \(C .\) $$ \begin{array}{l}{\mathbf{F}=(x+y) \mathbf{i}-\left(x^{2}+y^{2}\right) \mathbf{j}} \\ {C : \text { The triangle bounded by } y=0, x=1, \text { and } y=x}\end{array} $$

Step-by-Step Solution

Answer

Counterclockwise circulation is \(-\frac{7}{6}\) and outward flux is \(\frac{1}{6}\).

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 for a vector field \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j}\), \(\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\).

Step 2: Identify P and Q

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

Step 3: Compute Partial Derivatives

Calculate \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\). We have: \(\frac{\partial Q}{\partial x} = -2x\) and \(\frac{\partial P}{\partial y} = 1\).

Step 4: Set Up the Double Integral

Using Green's Theorem, set up the double integral for circulation: \[ \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \iint_R (-2x - 1) \, dA. \]

Step 5: Identify the Region R

Region \(R\) is the triangular region bounded by \(y=0\), \(x=1\), and \(y=x\). Describe the region using inequalities: \(0 \leq y \leq x\) and \(0 \leq x \leq 1\).

Step 6: Evaluate the Double Integral

Evaluate \[ \int_0^1 \int_0^x (-2x - 1) \, dy \, dx \]. The inner integral yields \(-2xy - y\) evaluated from \(y=0\) to \(y=x\), resulting in \(-2x^2 - x\). Integrate over \(x\) from 0 to 1: \[ \int_0^1 (-2x^2 - x) \, dx = \left[ -\frac{2}{3}x^3 - \frac{1}{2}x^2 \right]_0^1. \] This simplifies to \(-\frac{2}{3} - \frac{1}{2} = -\frac{7}{6}\).

Step 7: Find Outward Flux

Using Green's Theorem for the outward flux, set up \(\iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA \). Compute \(\frac{\partial P}{\partial x} = 1\) and \(\frac{\partial Q}{\partial y} = -2y\). The integral becomes \[ \iint_R (1 - 2y) \, dA \]. Evaluate \[ \int_0^1 \int_0^x (1 - 2y) \, dy \, dx \].

Step 8: Evaluate the Outward Flux Integral

Evaluate \[ \int_0^1 \left[ y - y^2 \right]_0^x \, dx = \int_0^1 (x - x^2) \, dx \]. This integral evaluates to \(\left[ \frac{1}{2}x^2 - \frac{1}{3}x^3 \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\).

Key Concepts

Line Integral

A line integral is a type of integral where integration is performed over a curve. It's akin to integrating a function along a path in space. In physics, this is often used to determine things like work done by a force field on a particle moving through the field.

In mathematics, if you have a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \), the line integral of \( \mathbf{F} \) along a curve \( C \) is given by the expression \( \oint_C \mathbf{F} \cdot d\mathbf{r} \). Here, \( d\mathbf{r} \) represents a tiny vector segment of the curve.

Line integrals are crucial in Green's Theorem. They help relate physical work or circulation around a boundary to more general properties of a region inside the boundary.

Vector Field

A vector field is a mathematical construction where a vector is associated with every point in a region of space. Imagine arrows placed across a field where each arrow has a specific length and direction based on given functions of coordinates like \( x \) and \( y \).

In the given exercise, the vector field is \( \mathbf{F} = (x+y) \mathbf{i} - (x^2+y^2) \mathbf{j} \). Here,\( (x+y) \) is the component along the \( x \)-axis, and \(- (x^2+y^2) \) is the component along the \( y \)-axis.

Vector fields are used extensively in physics to represent forces like electric or magnetic fields. In Green's Theorem, they are key, as they help determine how a line integral around a curve can be converted to a double integral over the area that the curve encloses.

Double Integral

A double integral extends the concept of an integral to functions of two variables. It is used to calculate areas, volumes, or other quantities over a two-dimensional region. Essentially, it's summing up values over a defined area.

In the context of Green's Theorem, a double integral is performed over the region \( R \) bounded by a closed curve \( C \). It's expressed as \( \iint_R f(x, y) \, dA \), where \( f(x, y) \) represents the function you wish to integrate over \( R \).

In the exercise, Green's Theorem transforms a line integral around the boundary of a triangle into a double integral over the triangle itself, simplifying calculations of circulation or flux through the triangular region.

Partial Derivatives

Partial derivatives deal with finding the derivatives of functions with multiple variables with respect to one variable while keeping others constant. They are essential in multivariable calculus.

In the given exercise, you have the vector field \( \mathbf{F} = (x+y) \mathbf{i} - (x^2 + y^2) \mathbf{j} \), implying \( P(x, y) = x + y \) and \( Q(x, y) = -(x^2 + y^2) \).

Using Green's Theorem requires finding partial derivatives like \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). For instance, \( \frac{\partial Q}{\partial x} = -2x \) and \( \frac{\partial P}{\partial y} = 1 \).

These calculations show how changes in one direction affect your vector field's behavior, aiding in transforming lines integrals to double integrals and vice versa through Green's Theorem.