Problem 165 - Calculus Volume 3

Question

Use Green’s theorem to evaluate \(\int_{C+}\left(y^{2}+x^{3}\right) d x+x^{4} d y,\) where \(C^{+}\) is the perimeter of square \([0,1] \times[0,1]\) oriented counterclockwise.

Step-by-Step Solution

Answer

The integral evaluates to 0.

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 D bounded by C. It states: \[ \oint_{C} (P \, dx + Q \, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]. Here, \( P(x, y) = y^2 + x^3 \) and \( Q(x, y) = x^4 \).

Step 2: Compute Partial Derivatives

Now, compute the partial derivatives: \( \frac{\partial Q}{\partial x} = 4x^3 \) and \( \frac{\partial P}{\partial y} = 2y \).

Step 3: Set Up the Double Integral

Insert the partial derivatives into Green's theorem: \[ \iint_{D} \left( 4x^3 - 2y \right) \, dA \]. Since D is the square \([0,1] \times [0,1]\), it is aligned with the coordinate axes.

Step 4: Evaluate the Double Integral

Evaluate the double integral over the region D: \[ \int_{0}^{1} \int_{0}^{1} (4x^3 - 2y) \, dx \, dy \]. First, integrate with respect to x: \[ \int_{0}^{1} (4x^3 - 2y) \, dx = \left. x^4 - 2yx \right|_{0}^{1} = 1 - 2y \].

Step 5: Integrate with Respect to y

Now, integrate the result with respect to y: \[ \int_{0}^{1} (1 - 2y) \, dy = \left. y - y^2 \right|_{0}^{1} = 1 - 1^2 = 0 \].

Key Concepts

Line Integral

The line integral is an integral where the function to be integrated is evaluated along a curve. In the context of Green's Theorem, a line integral calculates the circulation of a vector field around a simple closed curve \( C \). For the given exercise, the line integral is expressed as \( \oint_{C} (P \, dx + Q \, dy) \). Here, \( P(x, y) = y^2 + x^3 \) and \( Q(x, y) = x^4 \) represent the components of the vector field. This expression gives the total 'flow' or work done by the field, along the path or boundary \( C \).

Line integrals have crucial applications in physics and engineering, particularly for calculating work done by a force field along a path or current along a wire.

Key points about line integrals include:

• They are direction sensitive, meaning reversing the direction of \( C \) reverses the sign of the integral.
• The curve \( C \) can be piecewise smooth.

Green's Theorem provides an efficient way to convert this line integral into a double integral over the region \( D \) bounded by \( C \).

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one variable while keeping other variables constant. In the context of Green's Theorem, they are used to transform a line integral into a double integral.

To apply Green's Theorem, compute the partial derivatives:

• \( \frac{\partial Q}{\partial x} \) represents how the function \( Q \) changes as \( x \) changes, holding \( y \) constant. In this exercise, \( \frac{\partial Q}{\partial x} = 4x^3 \).
• \( \frac{\partial P}{\partial y} \) shows the change in \( P \) with respect to \( y \), when \( x \) is fixed. Here, \( \frac{\partial P}{\partial y} = 2y \).

These derivatives are then combined to form the integrand \( 4x^3 - 2y \) for the double integral over the region \( D \).

Understanding partial derivatives is essential for multi-variable calculus and helps solve complex problems like evaluating line integrals over curves.

Double Integral

A double integral is a way to compute the volume under a surface over a given region \( D \). In applying Green’s Theorem, the double integral \( \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA \) replaces the line integral over \( C \).

The evaluation of the double integral proceeds with respect to \( x \) followed by \( y \), over the entire region \( D \). For our exercise, this region is a square with sides from 0 to 1.

Steps to evaluate a double integral:

• Integrate the inner integral (with respect to \( x \), in our case), keeping \( y \) constant.
• Compute definite limits of the inner integral before integrating the outer integral (with respect to \( y \)).

For this exercise, these computations gave 0 as a result, due to the symmetry and cancellation of terms over the specified region.

Double integrals are powerful tools in calculating quantities that accumulate over two-dimensional regions. They appear in fields from physics (calculating mass, charge), to statistics (probability distributions).

Simple Closed Curve

A simple closed curve \( C \) is one that forms a loop without crossing itself, enclosing a region \( D \) in the plane. For Green's Theorem, this curve forms the boundary where the line integral is evaluated.

In the exercise, the simple closed curve is the perimeter of a square defined by \([0,1]\times[0,1]\), with a counterclockwise orientation. Such curves are essential for applying Green’s Theorem, which requires:

• The function inside the line integral is defined and continuous on and inside the curve \( C \).
• The curve \( C \) is oriented positively, meaning it bounds a region where, imagining walking along the curve, the region \( D \) is always on the left.

The properties of simple closed curves allow Green's Theorem to transform a line integral over \( C \) into a double integral over the enclosed region \( D \), simplifying many calculations.