Problem 159 - Calculus Volume 3

Question

Evaluate integral \(\oint_{C}\left(x^{2}+y^{2}\right) d x+2 x y d y,\) where \(\begin{array}{lllll}C & \text { is the curve that } & \text { follows } & \text { parabola }\end{array}\) \(y=x^{2}\) from \((0,0)(2,4),\) then the line from (2,4) to \((2,\) 0), and finally the line from (2,0) to (0,0) .

Step-by-Step Solution

Answer

The integral evaluates to 32.

Step 1: Parameterize the Curve Segments

We need to parameterize each segment of the curve \(C\):1. **Parabola from \((0,0)\) to \((2,4)\)**: Parameterize as \(x = t\) and \(y = t^2\) where \(t\) ranges from 0 to 2.2. **Line from \((2,4)\) to \((2,0)\)**: Since \(x\) is constant at 2, parameterize as \(x = 2\) and \(y = 4 - s\) where \(s\) ranges from 0 to 4.3. **Line from \((2,0)\) to \((0,0)\)**: Parameterize as \(x = 2 - t\) and \(y = 0\) where \(t\) ranges from 0 to 2.

Step 2: Calculate the Integral on Each Segment

Evaluate the line integral for each segment using the parameterizations:1. **Parabola Segment**: \[\int_{0}^{2} \left[(t^2 + t^4) \frac{dx}{dt} + 2t^3 \frac{dy}{dt}\right] dt = \int_{0}^{2} (t^2 + t^4 + 4t^4) dt \] Simplify and compute: \(\int_{0}^{2} (t^2 + 5t^4) dt = \left[\frac{t^3}{3} + t^5\right]_{0}^{2} = \frac{8}{3} + 32 = 96/3 + 8/3 = 104/3\)2. **Vertical Line Segment**: \[\int_{0}^{4} \left[4 - s\right] \cdot 0 + 2 \cdot 2 \cdot (4 - s) \frac{dy}{ds} ds = \int_{0}^{4} 4 \cdot 0 + 0 ds = 0\]3. **Horizontal Line Segment**: \[\int_{0}^{2} (2 - t)^2 \frac{dx}{dt} + 0 dt = \int_{0}^{2} -(2-t)^2 dt \] Perform the integration: \[-\int_{0}^{2} (4 - 4t + t^2) dt = -[4t - 2t^2 + \frac{t^3}{3}]_{0}^{2} = -[8 - 8 + \frac{8}{3}] = -\frac{8}{3}\]

Step 3: Sum the Results of Each Segment

Add the results of each segment to find the total value of the integral:- Parabola Segment: \(104/3\)- Vertical Line Segment: \(0\)- Horizontal Line Segment: \(-8/3\)Summing these, the total is \[(104/3) + 0 + (-8/3) = 96/3 = 32\]

Step 4: Final Step: Conclusion

The total value of the line integral over the entire curve \(C\) is \(32\). Thus, the integral evaluates to **32**.

Key Concepts

curve parameterization

Parameterizing a curve is an essential step in evaluating a line integral. It's the process of expressing the curve in terms of a single variable, usually denoted as \( t \), that varies over a specific interval. This transformation helps manage complex curves by breaking them down into simpler, linear expressions.

In the given exercise, the curve \( C \) consists of three segments, including a parabolic curve and two straight lines. Here's how each one is parameterized:

• Parabola from \((0,0)\) to \((2,4)\): Since the parabola follows the equation \( y = x^2 \), we can let \( x = t \) and naturally \( y = t^2 \). The parameter \( t \) then ranges from \( 0 \) to \( 2 \), covering the entire segment of the parabola.
• Vertical Line from \((2,4)\) to \((2,0)\): For the vertical line, \( x \) remains constant at \( 2 \). Here, \( y \) can be expressed as \( 4-s \), with \( s \) ranging from \( 0 \) to \( 4 \).
• Horizontal Line from \((2,0)\) to \((0,0)\): In this case, \( x \) changes from \( 2 \) to \( 0 \), parameterized as \( x = 2-t \). Meanwhile, \( y \) remains fixed at \( 0 \), with \( t \) running from \( 0 \) to \( 2 \).

By parameterizing each segment, it becomes much simpler to integrate over the entire curve.

vector calculus

Vector calculus is a mathematical field that extends calculus to vector fields, often used to compute aspects like line integrals. Line integrals are integrals over a curve, typically involving a vector field and capturing how a field behaves as you traverse a path.

In this exercise, the function to be integrated over the curve \( C \) is given by the differential expression \((x^2 + y^2) \, dx + 2xy \, dy\). Here,

• Vector Field Components: The vector field is \( \mathbf{F} = (x^2 + y^2, 2xy) \), representing how the field changes across the plane.
• Line Integral: The line integral along a curve collects the field's influence along that path. It's crucial for understanding fields' behavior in contexts like thermodynamics and electromagnetism.

Calculating the line integral involves substituting the parameterized forms \( dx = \frac{dx}{dt} \cdot dt \) and \( dy = \frac{dy}{dt} \cdot dt \) into the integral, turning it into a manageable form.

integral evaluation

Once a curve is parameterized, evaluating the integral is the next step. This is where calculus comes deeply into play, as it requires integrating expressions over specified intervals determined by the parameterization.

For our multi-segment curve, evaluate the integral separately for each segment using their unique parameterizations:

• Parabolic Segment: The evaluation of the integral here involves substituting the expressions for \( x \) and \( y \), resulting in an integral \( \int_{0}^{2} (t^2 + 5t^4) dt \). Calculating this provides the component's total area or field influence over its path.
• Vertical Line Segment: This segment turns out to contribute \( 0 \), as observed from \( \int_{0}^{4} 0 \, ds \). A zero result can occur when there's no net field engagement over a path segment.
• Horizontal Line Segment: Here, integrating \( -\int_{0}^{2} (4 - 4t + t^2) dt \) results in \( -\frac{8}{3} \), balancing out particular components of the curve.

Finally, summing the results from each path segment gives the overall integral of the curve \( C \). Completion of this step gives deep insight into the interplay of the curve and vector field in the context studied.