Problem 191 - Calculus Volume 3

Question

Use Green’s theorem to evaluate \(\oint_{C} x y d x+x^{3} y^{3} d y,\) where \(C\) is a triangle with vertices (0,0),(1,0) , and (1, 2) with positive orientation.

Step-by-Step Solution

Answer

The value of the integral is \( \frac{130}{21} \).

Step 1: Understand 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 states: \( \oint_{C} (P\,dx + Q\,dy) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). Here, \( P(x,y) = xy \) and \( Q(x,y) = x^3y^3 \).

Step 2: Calculate Partial Derivatives

Find the partial derivatives \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). For \( Q = x^3y^3 \), \( \frac{\partial Q}{\partial x} = 3x^2y^3 \). For \( P = xy \), \( \frac{\partial P}{\partial y} = x \).

Step 3: Set Up Double Integral Using Green's Theorem

Substitute the partial derivatives into Green’s Theorem: \( \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_{R} \left( 3x^2y^3 - x \right) \, dA \).

Step 4: Define Limits of Integration

The region \( R \) is a triangle with vertices \((0,0), (1,0), (1,2)\). Set the limits for \( x \) from 0 to 1. For a fixed \( x \), \( y \) ranges from 0 to the line \( y = 2x \).

Step 5: Evaluate the Double Integral

Compute the integral: \[ \int_{0}^{1} \int_{0}^{2x} (3x^2y^3 - x) \, dy \, dx \].First, integrate with respect to \( y \):\[ \int_{0}^{2x} (3x^2y^3 - x) \, dy = \left[ \frac{3x^2y^4}{4} - xy \right]_{0}^{2x} = \frac{3x^2(2x)^4}{4} - x(2x) = \frac{192x^6}{4} - 2x^2 = 48x^6 - 2x^2 \].Then integrate with respect to \( x \):\[ \int_{0}^{1} (48x^6 - 2x^2) \, dx = \left[ \frac{48x^7}{7} - \frac{2x^3}{3} \right]_{0}^{1} = \left( \frac{48}{7} - \frac{2}{3} \right) = \frac{144}{21} - \frac{14}{21} = \frac{130}{21} = \frac{130}{21} \].

Key Concepts

Line Integral

Line integrals are a fascinating concept in calculus and vector fields that allow us to integrate a function along a curve. Imagine you have a path or a curve, and you want to sum up some quantity (like work or heat transfer) along that path. That's where line integrals come in. For example, if you are walking up a hill, the line integral would take into account how steep the path is along your route and calculate the total work done.

The line integral of a vector field around a closed path is often involved in various applications like physics or engineering, indicating total circulation or flow around a curve. In our exercise involving Green's Theorem, the line integral around the triangle with vertices

• (0,0)
• (1,0)
• (1,2)

is transformed into a double integral over the region inside the triangle. This transformation simplifies the computation of complex integrals by shifting the focus from integrating over a path to an area.

Partial Derivatives

Partial derivatives are essential in multivariable calculus. They represent how a function changes as each variable is individually varied, while keeping others constant. When you have a function of multiple variables, partial derivatives help in understanding how the function behaves in any particular direction.

For example, given a function \( f(x, y) \), the partial derivative with respect to \( x \) (written as \( \frac{\partial f}{\partial x} \)) measures the change in \( f \) as \( x \) changes while \( y \) is held constant.

In the exercise, Green’s Theorem makes use of partial derivatives to convert a line integral into a double integral. We calculated:

• \( \frac{\partial Q}{\partial x} = 3x^2y^3 \)
• \( \frac{\partial P}{\partial y} = x \)

These derivatives help us express the vector field in terms of an area integral, a key step in applying Green’s Theorem.

Double Integral

A double integral is a method to integrate over a two-dimensional area. It's like finding the volume under a surface that is represented by a function \( f(x, y) \) over a specific region in the xy-plane.

For the exercise at hand, we use double integrals to find the value of the line integral over a closed curve, as stated by Green's Theorem. We calculated the double integral:\[ \iint_{R} \left( 3x^2y^3 - x \right) \, dA \]This gives a value that corresponds to the area integral of the vector field expressions, weighted by their partial derivatives. The process involves integrating first with respect to one variable, and then the other, covering the entire 2D region enclosed by the curve.

This method effectively captures the behavior of a 3D surface bounded by the curve, turning what could be a tough line integral into something far more manageable.

Limits of Integration

Determining the proper limits of integration is crucial when setting up a double integral. It ensures that we integrate over the correct region or area in the xy-plane.

In the given exercise, the region \( R \) is defined by a triangle. The vertices

• (0, 0)
• (1, 0)
• (1, 2)

outline the boundaries of this region. Thus, the limits for \( x \) range from 0 to 1. For each fixed \( x \), \( y \) ranges from 0 to the line given by \( y = 2x \).

These specific limits help maintain the proper orientation and size of the region that is being integrated over. Incorrect limits could lead to integrating over an entirely different region, producing wrong results. Recognizing these boundaries is often aided by sketching the region or visualizing its geometric constraints.