Problem 11 - Thomas Calculus

Question

Use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(\mathbf{F}\) and curve \(C .\) \(\mathbf{F}=(x-y) \mathbf{i}+(y-x) \mathbf{j}\) \(C :\) The square bounded by \(x=0, x=1, y=0, y=1\)

Step-by-Step Solution

Answer

The counterclockwise circulation is 0, and the outward flux is 2.

Step 1: Introduction to Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). It consists of two forms: the circulation form and the flux form. For circulation, if \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \), then the line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) is equal to the double integral \( \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).

Step 2: Identify P and Q

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

Step 3: Compute Partial Derivatives

Compute \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \):\( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y-x) = -1 \)\( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x-y) = -1 \)

Step 4: Calculate the Circulation Integral

Substitute the partial derivatives into the circulation form of Green's Theorem:\( \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_D (-1 + 1) \, dA = \iint_D 0 \, dA = 0 \)Thus, the circulation around \( C \) is 0.

Step 5: Introduction to the Flux Form

For the flux form of Green's Theorem, \( \oint_C \mathbf{F} \cdot d\mathbf{n} = \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA \).

Step 6: Calculate Partial Derivatives for Flux

Compute \( \frac{\partial P}{\partial x} \) and \( \frac{\partial Q}{\partial y} \):\( \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x-y) = 1 \)\( \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(y-x) = 1 \)

Step 7: Calculate the Flux Integral

Substitute the partial derivatives into the flux form of Green's Theorem:\( \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA = \iint_D (1 + 1) \, dA = \iint_D 2 \, dA \)The region \( D \) is a square with area 1 (since its sides are length 1), so \( \int_D 2 \, dA = 2 \times 1 = 2 \).Thus, the outward flux across \( C \) is 2.

Key Concepts

Vector Field

A vector field is a mathematical representation where a vector is assigned to each point in a space. This concept is essential in physical sciences to model forces like gravitational or electromagnetic fields.
In our exercise, the vector field is defined as \( \mathbf{F} = (x-y) \mathbf{i} + (y-x) \mathbf{j} \). Here, each vector in the field points in a direction determined by its components \( (x-y) \) and \( (y-x) \).

• \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors in the direction of the x-axis and y-axis, respectively.
• Understanding how vectors change across a region helps us study circulation and flux.

Line Integral

A line integral is used to evaluate a function along a curve, gathering information over a continuous path. In vector fields, it calculates quantities like work done by force fields over a path.
For Green's Theorem, the line integral of a vector field \( \mathbf{F} \) over a closed curve \( C \) helps establish relationships, converting a potentially complex line integral into a more manageable double integral over region \( D \) enclosed by \( C \).

• The line integral for circulation is expressed as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \).
• For flux, we consider \( \oint_C \mathbf{F} \cdot d\mathbf{n} \).

Circulation

Circulation of a vector field quantifies the tendency of the field to rotate around a defined path. This is crucial in understanding vortex-like movements for fields like wind or water currents.
By using the circulation form of Green's Theorem, the line integral around the curve \( C \) can be evaluated as a double integral. In our exercise, the expression \( \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \) goes to zero, showing the field has no net rotational effect over the region.

• This simplification through Green's Theorem saves computation time and effort.
• No rotation implies the "amount of twisting" is balanced over the boundary \( C \).

Flux

Flux measures the quantity of a field passing through a surface. In fluid mechanics and electromagnetism, flux is vital for analyzing systems where quantities flow across boundaries.
For our exercise, the flux is calculated through the flux form of Green's Theorem, \( \iint_D \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) \, dA \). The net result of 2 indicates a positive flow outwards across boundary \( C \).

• This integral accumulates all field lines crossing the boundary.
• Practical applications include understanding how fluids move through porous media or heat flow across surfaces.

Partial Derivatives

Partial derivatives measure how a function changes as each variable is varied independently. They are crucial in multivariable calculus, providing insight into the "slope" or change rate of functions with multiple inputs.
In our exercise, partial derivatives are used to evaluate the expressions in Green's Theorem:

• \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) are computed for circulation.
• \( \frac{\partial P}{\partial x} \) and \( \frac{\partial Q}{\partial y} \) are computed for flux calculations.

This "differentiation" reveals how each component of field \( \mathbf{F} \) shifts as \( x \) and \( y \) vary, acting as building blocks in understanding vector field dynamics.