Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use Green's Theorem to evaluate the given line integral. Begin by sketching the region \(S\). \(\oint_{C}\left(e^{3 x}+2 y\right) d x+\left(x^{2}+\sin y\right) d y\), where \(C\) is the rectangle with vertices \((2,1),(6,1),(6,4)\), and \((2,4)\).

Step-by-Step Solution

Answer

The value of the line integral is 72.

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. The theorem is given by: \( \oint_{C} M \, dx + N \, dy = \iint_{S} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( C \) is the positively oriented boundary of the region \( S \).

Step 2: Identify \( M \) and \( N \)

From the given line integral \( \oint_{C} (e^{3x}+2y) \, dx + (x^2 + \sin y) \, dy \), we identify \( M = e^{3x} + 2y \) and \( N = x^2 + \sin y \).

Step 3: Calculate Partial Derivatives

Calculate \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \).\( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2 + \sin y) = 2x \) \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{3x} + 2y) = 2 \).

Step 4: Set Up the Double Integral

According to Green's Theorem, \(\iint_{S} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA = \iint_{S} (2x - 2) \, dA \).The region \( S \) is the rectangle with vertices \((2,1), (6,1), (6,4), (2,4)\). Hence, the limits of integration are:- For \( x \): \( 2 \leq x \leq 6 \)- For \( y \): \( 1 \leq y \leq 4 \)

Step 5: Evaluate the Double Integral

Solve the double integral:\( \int_{1}^{4} \int_{2}^{6} (2x - 2) \, dx \, dy \)First, integrate with respect to \(x\):\( \int_{2}^{6} (2x - 2) \, dx = \left[ x^2 - 2x \right]_{2}^{6} = (36 - 12) - (4 - 4) = 24 \).Then, integrate with respect to \(y\):\( \int_{1}^{4} 24 \, dy = 24 \cdot [y]_{1}^{4} = 24 \cdot (4 - 1) = 72 \).

Key Concepts

Line Integral Basics

A line integral is a type of integral where the function to be integrated is evaluated along a curve or path. This concept is crucial in physics and engineering, particularly in fields like electromagnetism and fluid dynamics. Unlike a regular integral that sums values over an interval of a line, a line integral accumulates values along a curve. Think of it as walking along a path and adding up quantities like energy or mass as you go.

In our given problem, the line integral is represented as \( \oint_{C} (e^{3x} + 2y) \, dx + (x^2 + \sin y) \, dy \). Here, \( C \) is the boundary of the rectangle with specified vertices. To compute a line integral directly, one would parameterize the curve \( C \), break it into segments, and sum up contributions over each segment. However, using Green's Theorem simplifies the process by converting it into a more manageable double integral over the area bounded by \( C \). This is particularly useful for closed curves.

Understanding Double Integrals

Double integrals allow us to integrate functions over two-dimensional regions. They sum up tiny, infinitesimal contributions to form a total over an area, often denoted as a region \( S \). Think of them as evaluating the volume under a surface defined by a function, similar to how single integrals evaluate the area under a curve.

Using Green's Theorem, our original line integral around curve \( C \) is transformed into a double integral over the region \( S \) that \( C \) encloses. It becomes \( \iint_{S} (2x - 2) \, dA \), where \( dA \) is the differential area element that varies over the rectangle. The limits of integration are determined by the rectangle's boundaries: \( x \) varies from 2 to 6, and \( y \) varies from 1 to 4. Solving a double integral involves integrating first with respect to \( x \), then \( y \), which effectively simplifies our complex line integral process.

Role of Partial Derivatives

Partial derivatives measure how a function changes as one of its variables changes, holding other variables constant. They're a foundational concept in multivariable calculus, relevant in understanding the gradient, divergence, and curl of vector fields.

Green's Theorem uses partial derivatives to relate a line integral to a double integral. In our case, we found \( \frac{\partial N}{\partial x} = 2x \) and \( \frac{\partial M}{\partial y} = 2 \). These represent the rate of change of the functions \( N \, (x^2 + \sin y) \) and \( M \, (e^{3x} + 2y) \) in the \( x \) and \( y \) directions, respectively.

• \( \frac{\partial N}{\partial x} \) captures how \( N \) changes as \( x \) changes, while ignoring changes in \( y \).
• \( \frac{\partial M}{\partial y} \) captures how \( M \) changes as \( y \) changes, while ignoring changes in \( x \).

These derivatives are crucial because Green’s Theorem involves calculating the difference \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \), transforming the line integral into a solvable double integral. This highlights the interconnectedness of linear and area-based calculus methods.