Problem 186 - Calculus Volume 3

Question

Use Green's theorem to evaluate line integral \(\int_{C}(3 x-5 y) d x+(x-6 y) d y, \quad\) where \(\quad C \quad\) is ellipse \(\frac{x^{2}}{4}+y^{2}=1\) and is oriented in the counterclockwise direction.

Step-by-Step Solution

Answer

The value of the line integral is \( 12\pi \).

Step 1: Understanding Green's Theorem

Green's Theorem states that for a positively oriented, simple, closed curve \(C\) and a region \(D\) it encloses, for continuously differentiable functions \(M\) and \(N\) on an open region containing \(D\), \( \int_{C} (Mdx + Ndy) = \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \).

Step 2: Identify Functions and Derivatives

In our line integral, \( M(x,y) = 3x - 5y \) and \( N(x,y) = x - 6y \). The partial derivative of \(N\) with respect to \(x\) is \( \frac{\partial N}{\partial x} = 1 \), and the partial derivative of \(M\) with respect to \(y\) is \( \frac{\partial M}{\partial y} = -5 \).

Step 3: Compute Curl Component

Substitute the derivatives into the curl component of Green's Theorem: \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - (-5) = 6 \).

Step 4: Calculate the Double Integral over Region D

The region \(D\) is the ellipse \( \frac{x^2}{4} + y^2 = 1 \). The area of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( \pi ab \). Here, \( a = 2 \) and \( b = 1 \). So, the area \( A = \pi \cdot 2 \cdot 1 = 2\pi \).

Step 5: Evaluate the Final Integral

According to Green's Theorem, \( \iint_{D} 6 \, dA = 6 \cdot A \) where \(A = 2\pi\). This results in \( 6 \times 2\pi = 12\pi \).

Key Concepts

Line Integrals

A line integral is a way to integrate functions along a curve. Instead of focusing on a space or region, like with regular integrals, line integrals focus on the journey along a path. In simple terms, it adds up values of a function along this path.

In mathematical terms, if you have a curve \( C \), and functions \( M(x, y) \) and \( N(x, y) \), the line integral is written as:

• \( \int_C (Mdx + Ndy) \)

This notation essentially means you sum up tiny parts of each function, weighted by how much you move in the \( x \) and \( y \) directions along \( C \).

Line integrals are often used in physics to calculate work done by a force or circulation of a fluid over a path. In the context of our exercise, the line integral involves the functions \( 3x - 5y \) and \( x - 6y \), and the path is an ellipse. Green's Theorem assists in evaluating such line integrals by converting them into easier double integrals over the enclosed area.

Partial Derivatives

Partial derivatives are a core concept in calculus when dealing with functions of multiple variables. If a function depends on two or more variables, each of these variables can change, and thus they each have their own derivative.

For example, given a function \( f(x, y) \), where \( f \) depends on both \( x \) and \( y \), the partial derivative with respect to \( x \) (\( \frac{\partial f}{\partial x} \)) is the derivative that measures how \( f \) changes as \( x \) changes, keeping \( y \) constant.

In the exercise solution, the partial derivatives we calculated for \( N(x, y) = x - 6y \) and \( M(x, y) = 3x - 5y \) are crucial. Understanding what's happening at each point along the curve illustrates how the function changes over the surface.

• For \( N(x, y) \), the partial derivative with respect to \( x \) is \( \frac{\partial N}{\partial x} = 1 \).
• For \( M(x, y) \), the partial derivative with respect to \( y \) is \( \frac{\partial M}{\partial y} = -5 \).

Knowing these helps calculate the "curl" in Green's theorem, providing insights into how the function behaves over the surface of the region \( D \).

Ellipse

An ellipse is a geometric shape that looks like a stretched circle. It is defined mathematically as the set of points where the sum of the distances to two fixed points, called foci, is constant.

In standard form, the equation of an ellipse centered at the origin is given by:

• \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)

Here, \( a \) and \( b \) are the lengths of the semi-major and semi-minor axes, respectively. In this problem's context, the ellipse equation becomes \( \frac{x^2}{4} + y^2 = 1 \), suggesting \( a = 2 \) and \( b = 1 \).

The area inside an ellipse is calculated using the formula \( \pi ab \), which is similar to finding the area of a circle. However, for ellipses, since they are stretched, we multiply \( \pi \) by both axes' lengths.

The ellipse's area in our exercise is calculated as \( 2\pi \). This area is essential because Green's Theorem uses the area to transform our line integral into a double integral, simplifying our evaluation for curves like the ellipse.