Problem 6 - 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 \(C .\) $$ \begin{array}{l}{\mathbf{F}=\left(x^{2}+4 y\right) \mathbf{i}+\left(x+y^{2}\right) \mathbf{j}} \\ {C : \text { The square bounded by } x=0, x=1, y=0, y=1}\end{array} $$

Step-by-Step Solution

Answer

Circulation: -3; Outward Flux: 2.

Step 1: Understand Green's Theorem

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

Step 2: Identify Components of \( \mathbf{F} \) and Curve \( C \)

The given vector field is \( \mathbf{F} = (x^2 + 4y) \mathbf{i} + (x + y^2) \mathbf{j} \). Here, \( P(x, y) = x^2 + 4y \) and \( Q(x, y) = x + y^2 \). The curve \( C \) is the boundary of the square defined by \( x = 0 \), \( x = 1 \), \( y = 0 \), and \( y = 1 \).

Step 3: Compute Partial Derivatives

Calculate the partial derivative of \( Q \) with respect to \( x \): \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x + y^2) = 1 \). Then, calculate the partial derivative of \( P \) with respect to \( y \): \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2 + 4y) = 4 \).

Step 4: Evaluate the Double Integral

Using Green's Theorem, we have to compute: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dA = \iint_R (1 - 4) \, dA = \iint_R (-3) \, dA \] The region \( R \) is the unit square \( [0, 1] \times [0, 1] \), so: \[ \iint_R (-3) \, dA = \int_0^1 \int_0^1 (-3) \, dx \, dy = -3 \times (1) = -3 \]

Step 5: Calculate the Outward Flux

The outward flux is calculated using another form of Green's Theorem for the vector field out of the curve. Using \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right) \, dA \), where \( \mathbf{n} \) is the outward unit normal. Compute the derivatives: \( \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x^2 + 4y) = 2x \) and \( \frac{\partial Q}{\partial y} = 2y \). Thus: \[ \iint_R (2x + 2y) \, dA = \int_0^1 \int_0^1 (2x + 2y) \, dx \, dy \] Evaluate integrals: \[ = \int_0^1 \left[ x^2 + 2yx \right]_0^1 \, dy = \int_0^1 (1 + 2y) \, dy = \left[ y + y^2 \right]_0^1 = 1 + 1 = 2 \]

Step 6: Final Answer

The counterclockwise circulation of \( \mathbf{F} \) around \( C \) is \(-3\) and the outward flux of \( \mathbf{F} \) across \( C \) is \(2\).

Key Concepts

Vector Field

A vector field is a mathematical function that assigns a vector to every point in a specific domain, typically in two or three-dimensional space. In simple terms, imagine a map with arrows attached to every point, indicating direction and magnitude. These vectors might represent various physical quantities, such as velocity in fluid dynamics or force fields in electromagnetism. Here, the vector field given by

• \( \mathbf{F} = (x^2 + 4y) \mathbf{i} + (x + y^2) \mathbf{j} \)

specifies that at every point

• \((x, y)\)

, the vector is calculated using the functions

• \( x^2 + 4y \)
• \( x + y^2 \)

for its horizontal and vertical components respectively. Understanding a vector field helps to analyze how various forces or velocities operate over multi-dimensional spaces.

Line Integral

A line integral, often denoted as

• \( \oint_C \mathbf{F} \cdot d\mathbf{r} \)

, is a way to integrate a function over a curve. It's particularly useful in physics and engineering for calculating work done by a force field when moving along a curve. Consider walking along a path while wind blows against you; the line integral calculates the total effect of the wind on your walk.
For the given vector field

• \( \mathbf{F} \)

, Green’s Theorem allows us to convert this line integral over the boundary curve

• \( C \)

into a double integral over the region enclosed by the curve. This powerful tool greatly simplifies calculations when evaluating circulation and flux.

Double Integral

A double integral is a way to integrate over a two-dimensional area. With a double integral, we can compute total quantities over regions of a plane. It's critical in calculating areas, volumes, and in context here, relates to integrating vector fields over regions.
To compute the double integral in Green's Theorem:

• \[ \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]

we evaluate over region

• \( R \)

using the partial derivatives of the vector field components. This simplifies the task of finding circulation or flux, transforming a line integral problem into a problem of computing a double integral over the plane area enclosed by

• \( C \)

.

Partial Derivatives

Partial derivatives measure how a function changes as one particular variable changes, keeping others constant. In vector fields, they reveal local rates of change with respect to each variable direction. For a function

• \( f(x, y) \)

, the partial derivative with respect to

• \( x \)

, denoted

• \( \frac{\partial f}{\partial x} \)

, shows how

• \( f \)

changes if

• \( x \)

changes but

• \( y \)

does not.
For Green’s Theorem, understanding the partial derivatives

• \( \frac{\partial Q}{\partial x} \)
• \( \frac{\partial P}{\partial y} \)

is essential. They are used to convert line integrals into double integrals, transforming complex boundary problems into simpler region-based calculations.