Problem 198 - Calculus Volume 3

Question

Use Green's theorem in a plane to evaluate line integral \(\oint_{C}\left(x y+y^{2}\right) d x+x^{2} d y, \quad\) where \(C\) is a closed curve of a region bounded by \(y=x\) and \(y=x^{2}\) oriented in the counterclockwise direction.

Step-by-Step Solution

Answer

The line integral evaluates to \( \frac{1}{12} \).

Step 1: State Green's Theorem

Green's Theorem relates a line integral around a simple closed curve to a double integral over the region it encloses. The theorem is given by \( \oint_C (M\, dx + N\, dy) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( M(x, y) = x y + y^2 \) and \( N(x, y) = x^2 \).

Step 2: Find Partial Derivatives

Determine \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \):- \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (x^2) = 2x \).- \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (x y + y^2) = x + 2y \).

Step 3: Set Up the Double Integral

The double integral over the region \( R \) becomes\[ \iint_R (2x - (x + 2y)) \, dA = \iint_R (x - 2y) \, dA \].

Step 4: Determine the Limits of Integration

The region \( R \) is bounded by the curves \( y = x^2 \) and \( y = x \). For the given curves, x ranges from 0 to 1 (the points where they intersect). For a fixed x, y goes from \( x^2 \) to \( x \). This gives the limits:- For x: 0 to 1- For y: \( x^2 \) to \( x \)

Step 5: Evaluate the Double Integral

Evaluate the integral:\[ \int_{0}^{1} \int_{x^2}^{x} (x - 2y) \, dy \, dx \]First, integrate with respect to y:\[ \int_{x^2}^{x} (x - 2y) \, dy = \left[ xy - y^2 \right]_{x^2}^x \]\[ = (x^2 - x^4) - (x^3 - x^4) = x^2 - x^3 \]Then integrate with respect to x:\[ \int_{0}^{1} (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \]

Step 6: Conclude the Solution

According to Green's theorem, the value of the line integral \( \oint_{C}(x y + y^{2}) \, dx + x^{2} \, dy \) is given by the double integral, which evaluates to \( \frac{1}{12} \).

Key Concepts

Line Integrals

A line integral is a generalization of an integral that helps in evaluating the accumulation of a field along a curve. In simpler terms, it measures how much a vector field contributes as you move along a path. The concept of a line integral is widely used in physics and engineering, especially in the calculation of work done by a force field.

When dealing with line integrals of vector fields, we often use the notation \( \oint_C Mdx + Ndy \), where \( C \) represents a closed curve or path. In our original exercise, this was presented through the curve \( C \) defined by the boundaries \( y = x \) and \( y = x^2 \).

• The path needs to be oriented, usually in a counterclockwise manner to comply with Green's Theorem conditions.
• The integral calculates the "flow" of the field around the boundary of the region.

Here, understanding line integrals sets the stage for applying Green’s Theorem, as we did in converting the line integral into an easier double integral.

Double Integrals

Double integrals extend the concept of integration to two dimensions. They allow us to calculate the volume under a surface or accumulate quantities over a given area in a plane.

In Green’s Theorem, we convert the line integral into a double integral over the region it encloses, which is often simpler to evaluate. For our original problem, after applying Green’s Theorem, the expression \( \iint_R (2x - (x+2y)) \, dA \) was set inside the region bounded by \( y = x \) and \( y = x^2 \).

• Double integrals commonly require choosing appropriate limits of integration that represent the region \( R \).
• In polar, Cartesian, or other coordinates—the limits reflect the bounding curves or lines.

For step-by-step solving, the integration in one variable is first done before moving to the next, as demonstrated when we calculate \( \iint_R (x - 2y) \, dA \). This gives us the solution \( \frac{1}{12} \), which simplifies the complexity involved in analyzing each line segment individually.

Partial Derivatives

Partial derivatives are a crucial part of calculus for dealing with functions of multiple variables. They represent how a function changes as each variable is altered while keeping others constant.

In the context of Green's Theorem and our example, partial derivatives help translate a vector field into a divergence or curl form needed for integration over a region. For the original task:

• We found \( \frac{\partial N}{\partial x} = 2x \) from \( N(x,y) = x^2 \).
• And \( \frac{\partial M}{\partial y} = x + 2y \) from \( M(x,y) = xy + y^2 \).

These calculations are essential since Green's Theorem involves the expression \( \iint_R \left ( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right ) \, dA \). Understanding partial derivatives enables us to manipulate these expressions effectively, ensuring a smooth transition from line integrals to double integrals, thereby simplifying our computations.