Question

\(\oint(2 y d x-3 x d y)\) around the square bounded by \(x=3, x=5, y=1\) and \(y=3\)

Short answer

The closed path integral is 0.

Step-by-step solution

Identify the boundaries

The square is bounded by the lines: 1. x = 3 2. x = 5 3. y = 1 4. y = 3

Parameterize the boundary

The boundary of the square can be parameterized in four segments:Segment 1: From (3, 1) to (5, 1): ewline ewline Segment 2: From (5, 1) to (5, 3);Segment 3: From (5, 3) to (3, 3);Segment 4: From (3, 3) to (3, 1) .

Evaluate the integral on segment 1

For segment 1: From (3, 1) to (5, 1): Here, y = 1, so dy = 0:ewline Evaluate ewline ewline ewline

Evaluate the integral on segment 2

For segment 2: From (5, 1) to (5, 3): Here, x = 5, so dx = 0: Evaluate

Evaluate the integral on segment 3

For ségment 3: From (5, 3) to (3, 3): Here, y = 3, so dy = 0:Evaluate .

Evaluate the integral on segment 4

For segment 4: From (3, 3) to (3, 1): Here, x = 3, so dx = 0: ewline Evaluate.

Sum all the integrals

Sum up all the integral results from segments 1 to 4 to find the total integral.

Key concepts

parameterization

Parameterization involves describing a curve using a parameter, often denoted as 't'. This is essential in line integrals because it simplifies the function you're integrating over. By parameterizing each segment of the boundary, you can easily compute the individual integrals. For instance, in our exercise, we parameterized the boundary of the square into four segments.

• Segment 1: \(x = t, y = 1 \) where \(3 \leq t \leq 5\)
• Segment 2: \(x = 5, y = t \) where \(1 \leq t \leq 3\)
• Segment 3: \(x = t, y = 3 \) where \(5 \geq t \geq 3\)
• Segment 4: \(x = 3, y = t \) where \(3 \geq t \geq 1\)

By using parameterization, we converted the boundaries into functional forms that are easier to work with.

Green's Theorem

Green's Theorem links a line integral around a simple curve to a double integral over the plane region bounded by the curve. This theorem is particularly useful for evaluating line integrals. It states:
\[ \oint_C (L dx + M dy) = \iint_R \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA \]
For our problem, \(L = 2y\) and \(M = -3x\). Calculating \(\frac{\partial M}{\partial x} = -3\) and \(\frac{\partial L}{\partial y} = 2\), we find that:
\[ \oint (2 y dx - 3 x dy) = \iint_R (-3 - 2) dA = -5 \iint_R dA \] The region \(R\) is a square with area \(4 \text{ square units} (2 \times 2)\). So, the result is \(-5 \times 4 = -20\). This simplified approach using Green's Theorem helps us avoid computing multiple individual line integrals.

multivariable calculus

Multivariable calculus extends calculus to functions of several variables. It includes finding and interpreting partial derivatives, multiple integrals, and vector calculus. Concepts like line integrals belong to this field and involve integrating along a path in a vector field.

• **Partial Derivatives:** These indicate how a function changes as one of the variables is varied, holding others constant.
• **Multiple Integrals:** These are integrals over higher-dimensional spaces, like double and triple integrals.
• **Vector Calculus:** This deals with vector fields and operations on them, such as divergence and curl.

In our exercise, we use multivariable calculus to evaluate the line integral around a square path, illustrating how several concepts come together to solve a problem.

vector fields

A vector field assigns a vector to each point in a subset of space. In two dimensions, this is given by \((L(x, y), M(x, y))\). These vectors can represent various phenomena like fluid flow or electromagnetic fields. Understanding vector fields is crucial for line integrals as it shows how vectors vary along a path.
In our exercise, the vector field is given by \((2y, -3x)\). The components of this vector field are functions of the position \((x, y)\). When computing the line integral over this vector field, we measure how these vectors interact with the path defined by the boundaries of the square.
Remember the key points:

• **Line Integrals:** Measure the cumulative effect of the vector field along a path.
• **Divergence and Curl:** Indicate how vectors diverge from or rotate around a point, helping us understand the field's behavior.

Hence, vector fields are foundational to grasping and calculating line integrals in multivariable calculus.