Problem 8 - Advanced Engineering Mathematics

Question

Use Green's theorem to evaluate the given line integral. $$ \begin{aligned} &\oint_{C}(x-3 y) d x+(4 x+y) d y \text { , where } C \text { is the rectangle with }\\\ &\text { vertices }(-2,0),(3,0),(3,2),(-2,2) \end{aligned} $$

Step-by-Step Solution

Answer

The line integral evaluates to 70.

Step 1: Determine the Vector Field Components

Identify the vector field components from the line integral. The integrand is given as \((x - 3y)\, dx + (4x + y)\, dy\). From this, the components of the vector field \(\mathbf{F}\) are \(P = x - 3y\) and \(Q = 4x + y\).

Step 2: Write Green's Theorem Formula

Green's theorem states that for a positively oriented, simple closed curve \(C\) and the region \(R\) it encloses, \(\oint_C P\, dx + Q\, dy = \int \int_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\, dA\).

Step 3: Compute Partial Derivatives

Find \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\). Calculate \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4x + y) = 4\) and \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x - 3y) = -3\).

Step 4: Set Up the Double Integral

Insert the partial derivatives into Green's Theorem formula: \(\int \int_R (4 - (-3))\, dA = \int \int_R 7\, dA\).

Step 5: Define the Region of Integration

The region \(R\) is the rectangle defined by the vertices \((-2,0), (3,0), (3,2), (-2,2)\). Thus, the integration limits are \(-2 \leq x \leq 3\) and \(0 \leq y \leq 2\).

Step 6: Evaluate the Double Integral

Compute \(\int_{x=-2}^{3} \int_{y=0}^{2} 7\, dy\, dx\). First, integrate with respect to \(y\): \(\int_{0}^{2} 7\, dy = 7y\big|_{0}^{2} = 14\). Then integrate the result with respect to \(x\): \(\int_{-2}^{3} 14\, dx = 14x\big|_{-2}^{3} = 14(3) - 14(-2) = 42 + 28 = 70\).

Step 7: Conclude with the Result

The value of the line integral \(\oint_{C}(x-3 y) d x+(4 x+y) d y\) using Green's theorem is 70.

Key Concepts

Line Integral

A line integral is a way to integrate a function along a curve, capturing the notion of adding up quantities along a path. This is especially useful in physics and engineering, where you may need to calculate work done by a force field in moving an object along a certain path.
In our problem, the line integral is \[ \oint_{C}(x-3 y) \; dx +(4 x+y) \; dy \] where \(C\) is the closed path along the rectangle with specified vertices. The purpose of using Green's Theorem is to convert this line integral into an easier-to-evaluate double integral.

Vector Field

A vector field assigns a vector to each point in space, and in line integrals, it represents a force field. In our context, we identified the vector field \( \mathbf{F} \) from the line integral as having components:

• \( P(x, y) = x - 3y \)
• \( Q(x, y) = 4x + y \)

This means at every point \( (x, y) \) in space, \( \mathbf{F} \) gives a vector that's calculated using these components, depicting how the field behaves at that point.

Double Integral

A double integral allows us to compute the accumulation of a quantity over a two-dimensional area. By applying Green's theorem, we convert the line integral into a double integral which is often easier to evaluate. The formula given by Green's theorem is:\[ \oint_C P \; dx + Q \; dy = \int \int_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \; dA \]In our problem, this became:\[ \int \int_R 7 \; dA \]
This double integral is evaluated over the rectangular region \(R\), making it straightforward to calculate using rectangular coordinates.

Partial Derivatives

Partial derivatives are used to measure how a function changes as only one of the variables changes while others are held constant. In Green's theorem, partial derivatives help in transforming a line integral into a double integral. From our components:

• The partial derivative \( \frac{\partial Q}{\partial x} \) tells us how \( Q \) changes with \( x \), and in this case, it is 4.
• \( \frac{\partial P}{\partial y} \) shows how \( P \) changes with \( y \), which is -3.

These partial derivatives combined in Green’s theorem give us the term \( 4 - (-3) = 7 \), crucial for setting up our double integral for evaluation.