Problem 132 - Calculus Volume 3

Question

For the following exercises, $$ \mathbf{F}(x, y)=2 x y^{2} \mathbf{i}+\left(2 y x^{2}+2 y\right) \mathbf{j} $$ \(G(x, y)=(y+x) \mathbf{i}+(y-x) \mathbf{j}, \quad\) and let \(C_{1}\) be the curve consisting of the circle of radius 2 , centered at the origin and oriented counterclockwise, and \(C_{2}\) be the curve consisting of a line segment from (0,0) to (1,1) followed by a line segment from (1,1) to (3,1) . Calculate the line integral of \(\mathbf{F}\) over \(\mathrm{C}_{2}\).

Step-by-Step Solution

Answer

The line integral over \( C_2 \) is \( \frac{19}{2} \).

Step 1: Parametrize the First Segment of C2

The first segment of \(C_2\) is a line from \((0,0)\) to \((1,1)\). We parametrize this line segment by \( \mathbf{r}_1(t) = (t, t) \) for \( t \in [0, 1] \).

Step 2: Compute the Differential for the First Segment

For the parametrization \( \mathbf{r}_1(t) = (t, t) \), the differential \( d\mathbf{r}_1 \) is \( (1, 1) \, dt \).

Step 3: Evaluate F on the First Segment

Substitute \( x = t \) and \( y = t \) into \( \mathbf{F}(x, y) = 2xy^2 \mathbf{i} + (2yx^2 + 2y) \mathbf{j} \) to get \( \mathbf{F}(t, t) = 2t^3 \mathbf{i} + (2t^3 + 2t) \mathbf{j} \).

Step 4: Compute the Line Integral Over the First Segment

The integral \( \int_{C_{2,1}} \mathbf{F} \cdot d\mathbf{r}_1 \) becomes \( \int_0^1 [(2t^3)(1) + (2t^3 + 2t)(1)] \, dt = \int_0^1 (4t^3 + 2t) \, dt \). Calculate the integral to find \( \left[ t^4 + t^2 \right]_0^1 = 1 + \frac{1}{2} = \frac{3}{2} \).

Step 5: Parametrize the Second Segment of C2

The second segment is from \((1,1)\) to \((3,1)\). Parametrize this line segment with \( \mathbf{r}_2(t) = (1 + 2t, 1) \), where \( t \in [0, 1] \).

Step 6: Compute the Differential for the Second Segment

For \( \mathbf{r}_2(t) = (1 + 2t, 1) \), the differential \( d\mathbf{r}_2 \) is \( (2, 0) \, dt \).

Step 7: Evaluate F on the Second Segment

For the second segment, substitute \( x = 1 + 2t \) and \( y = 1 \) into \( \mathbf{F}(x, y) = 2xy^2 \mathbf{i} + (2yx^2 + 2y) \mathbf{j} \), resulting in \( \mathbf{F}(1+2t, 1) = 2(1+2t) \mathbf{i} + (2(1)(1+2t)^2 + 2) \mathbf{j} \). This simplifies to \( (2 + 4t) \mathbf{i} + (2(1+4t+4t^2) + 2) \mathbf{j} = (2 + 4t) \mathbf{i} + (4 + 8t + 8t^2) \mathbf{j} \).

Step 8: Compute the Line Integral Over the Second Segment

The integral \( \int_{C_{2,2}} \mathbf{F} \cdot d\mathbf{r}_2 \) becomes \( \int_0^1 [(2 + 4t)(2) + (4 + 8t + 8t^2)(0)] \, dt = \int_0^1 (4 + 8t) \, dt \). Calculate the integral to find \( \left[ 4t + 4t^2 \right]_0^1 = 4 + 4 = 8 \).

Step 9: Add the Line Integrals of Both Segments

The line integral over the entire curve \( C_2 \) is the sum of the integrals over each segment: \( \frac{3}{2} + 8 = \frac{19}{2} \).

Key Concepts

Vector Fields

Vector fields provide an essential way of representing various physical quantities as functions of position in space. They are crucial in many physics and engineering applications, for example in describing magnetic fields or fluid flow. A vector field assigns a vector to every point in a subset of space. In mathematical language, if you have a vector field \( \mathbf{F}(x, y) = 2xy^2 \mathbf{i} + (2yx^2 + 2y) \mathbf{j} \), this field means for each point \( (x, y) \), there is a vector pointing to a specific direction and having a set magnitude.
Understanding vector fields helps in visualizing how a quantity like force progresses through space. You are calculating line integrals over these fields to understand how these forces work along a specific path or curve.

Curve Parametrization

Parametrization is a technique used to represent a curve by expressing its coordinates as functions of a variable, usually denoted by \( t \), which acts as a parameter.
When parametrizing a curve, each point on the curve is associated with a specific value of \( t \). For the first segment of curve \( C_2 \), we used \( \mathbf{r}_1(t) = (t, t) \) where \( t \) spans from 0 to 1. It means as \( t \) varies from 0 to 1, the curve starts at \( (0,0) \) and ends at \( (1,1) \).
Similar steps are followed for the second segment, using the parameter \( \mathbf{r}_2(t) = (1 + 2t, 1) \) from \( t = 0 \) to \( t = 1 \), which results in a movement from \( (1,1) \) to \( (3,1) \). This process provides a standardized way to compute expressions along the curve.

Differential Equations

In the context of line integrals, differentials play a crucial role. A differential equation involves functions and their derivatives, describing how these functions change. While parametric differentials like \( d\mathbf{r}(t) = (1, 1) \, dt \) suggest how small changes in \( t \) translate to changes along the path.
By finding how a vector field interacts with these differentials, we calculate the work done by the field or any other integral quantity along the path. This reveals important insights about the system's behavior. In essence, using differentials helps to integrate vector fields over curves effectively.

Calculus of Vector Functions

The calculus of vector functions involves integrating or differentiating functions that have vector outputs. For line integrals, the goal is to compute an integral involving a dot product between vector fields and the derivative of a curve's parametrization.
This ensures that the vector field contribution is measured correctly along every infinitesimal part of the curve. Evaluating \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \), as in our example, involves lining up the field and the curve's direction and magnitude to sum contributions accurately.
The final result gives us the total accumulation of the vector field's influence along the path, which is crucial for applications in physics and engineering.