Problem 30 - Thomas Calculus

Question

Integral dependent only on area Show that the value of $$ \oint_{C} x y^{2} d x+\left(x^{2} y+2 x\right) d y $$ around any square depends only on the area of the square and not on its location in the plane.

Step-by-Step Solution

Answer

The integral evaluates to twice the area of the square and depends only on the square's area.

Step 1: Understand the Problem

We need to evaluate the line integral \(\oint_{C} x y^{2} \, dx + (x^{2} y + 2x) \, dy\) around a square and show that it depends only on the area of the square, not its position. This suggests using Green's Theorem, which relates a line integral around a closed curve to a double integral over the region it encloses.

Step 2: Apply Green's Theorem

Green's Theorem states \[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA, \] where \(P(x, y) = x y^2\) and \(Q(x, y) = x^2 y + 2x\). We need to find \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\).

Step 3: Calculate Partial Derivatives

Compute \(\frac{\partial Q}{\partial x}\): \[ \frac{\partial}{\partial x}(x^2 y + 2x) = 2xy + 2. \]Compute \(\frac{\partial P}{\partial y}\): \[ \frac{\partial}{\partial y}(x y^2) = 2xy. \]

Step 4: Substitute into Green's Theorem

Substitute the partial derivatives into Green's Theorem expression:\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (2xy + 2) - 2xy = 2. \]Thus, the integral simplifies to \(\iint_{R} 2 \, dA\).

Step 5: Evaluate the Double Integral

The double integral \(\iint_{R} 2 \, dA\) over the region \(R\) simply equals 2 times the area of the region \(R\). If \(R\) is a square with area \(A\), then \[ \iint_{R} 2 \, dA = 2 \cdot A. \]

Step 6: Conclude the Integral's Dependence

The line integral's value, given by \(2A\), depends only on the area \(A\) of the square and factors out any dependence on the square's location within the plane.

Key Concepts

Green's Theorem

Green's Theorem is a fundamental concept in vector calculus that establishes a relationship between a line integral around a simple closed curve and a double integral over the plane region bounded by that curve. The theorem states:

• \( \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \)

where:

• \(C\) is the closed curve,
• \(R\) is the region it encloses,
• \(P(x, y)\) and \(Q(x, y)\) are continuously differentiable functions.

The beauty of Green's Theorem lies in its ability to convert the problem of calculating a potentially complex line integral into a potentially simpler double integral. This is particularly useful when the double integral is easier to evaluate over the specified region. In the context of the exercise, applying Green's Theorem allows us to understand how the value of the line integral solely depends on the area of the square rather than its specific location.

Partial Derivatives

Partial derivatives are used in calculus to examine the rate at which a function changes with respect to one of its variables, keeping the other variables constant. This concept is crucial in several fields, including physics, engineering, and, as we've seen in our solution, mathematics.

• For a function \(f(x, y)\), the partial derivative with respect to \(x\) is denoted by \(\frac{\partial f}{\partial x}\) and represents how \(f\) changes as \(x\) changes while \(y\) remains constant.
• Similarly, \(\frac{\partial f}{\partial y}\) denotes the rate of change of \(f\) with respect to \(y\) keeping \(x\) constant.

In the solution for the exercise, we computed partial derivatives for functions \(Q(x, y) = x^2 y + 2x\) and \(P(x, y) = xy^2\) to apply Green's Theorem:

• \( \frac{\partial Q}{\partial x} = 2xy + 2 \)
• \( \frac{\partial P}{\partial y} = 2xy \)

These derivatives were essential to find the expression \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\) needed in Green's Theorem.

Double Integrals

Double integrals extend the concept of integration to functions of two variables, providing a way to calculate the volume under a surface over a given region. When solving the exercise, we ultimately transformed the line integral into a double integral,

• \(\iint_{R} 2 \; dA\)

This means we are calculating the total of the function 2 over the area \(A\) of the square.

• The double integral \(\iint_{R} 2 \; dA = 2 \cdot \) area \(A\)

is straightforward to evaluate, as it simplifies to a constant multiplied by the area of the region. Thus, it demonstrates that the original line integral depends only on this area, revealing that its location does not affect its value. This illustrates the power and simplicity of converting a problem with possibly variable-dependent complexity into a well-defined geometric interpretation, as facilitated using Green's Theorem.