Problem 133 - 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 \(\mathbf{G}\) over \(C_{2}\).

Step-by-Step Solution

Answer

The line integral of \(\mathbf{G}\) over \(C_2\) is 7.

Step 1: Parameterize the First Segment of C2

The first segment of \(C_2\) is a line segment from \((0,0)\) to \((1,1)\). To parametrize, let \(x = t\) and \(y = t\) for \(t\) in \([0,1]\). Thus, we have the parametrization \((x,y) = (t,t)\) with \(0 \leq t \leq 1\).

Step 2: Calculate the Line Integral Over the First Segment of C2

For the line integral of \(\mathbf{G}\) along the first segment, \(\int_{C_2} \mathbf{G} \cdot d\mathbf{r}\), we compute the dot product:\[\mathbf{G}(t,t) = (t+t) \mathbf{i} + (t-t) \mathbf{j} = (2t,0)\]The derivative \(d\mathbf{r} = (dx,dy) = (dt,dt)\). So, the line integral becomes: \[\int_{0}^{1} (2t,0) \cdot (dt,dt) = \int_{0}^{1} 2t \cdot dt = \int_{0}^{1} 2t \, dt \].Integrating this gives \[t^2\] evaluated from 0 to 1:\[\left[t^2\right]_0^1 = 1^2 - 0^2 = 1\].

Step 3: Parameterize the Second Segment of C2

The second segment of \(C_2\) is a line segment from \((1,1)\) to \((3,1)\). Parametrize it by letting \(x = 1 + 2s\) and \(y = 1\) for \(s\) in \([0,1]\). Hence, the parametrization is \((x,y) = (1+2s, 1)\) with \(0 \leq s \leq 1\).

Step 4: Calculate the Line Integral Over the Second Segment of C2

For the second segment's integral, calculate the dot product:\[\mathbf{G}(1+2s,1) = (1+2s+1) \mathbf{i} + (1-(1+2s)) \mathbf{j} = (2+2s, -2s)\]The derivative \(d\mathbf{r} = (dx,dy) = (2ds, 0)\).Compute the line integral:\[\int_{0}^{1} (2+2s, -2s) \cdot (2ds,0) = \int_{0}^{1} 2(2+2s) ds\].This simplifies to \[\int_{0}^{1} 4 + 4s \, ds\], which integrates to \[4s + 2s^2\] evaluated from 0 to 1:\[\left[4s + 2s^2\right]_0^1 = 4(1) + 2(1)^2 = 6\].

Step 5: Combine Results from Both Segments of C2

The total line integral over \(C_2\) is the sum of the integrals over both segments:\[ \int_{C_2} \mathbf{G} \cdot d\mathbf{r} = 1 + 6 = 7 \].

Key Concepts

Vector Fields

Imagine a field made up of arrows pointing in various directions, each with a different length and direction. This is essentially what a vector field is—a function that assigns a vector to every point in a space. In mathematics, vector fields often represent velocity fields of fluid flow or force fields in physics.

Vectors are represented by components along specified directions. For instance, the vector field \( \mathbf{G}(x, y) = (y+x) \mathbf{i} + (y-x) \mathbf{j} \) means each point \((x, y)\) corresponds to a vector in this 2D field. The components \((y+x)\) and \((y-x)\) dictate the horizontal and vertical directions, respectively. Understanding vector fields helps visualizing the forces or velocities at play, which is crucial when calculating line integrals.

Parametrization of Curves

Parametrization is a fancy term for describing curves in a more manageable way using parameters. Instead of dealing with \(x\) and \(y\) values directly, we represent them in terms of another variable, often \(t\) or \(s\).

For instance, consider a curve described by the line segments in solution steps. The first segment of \(C_2\) is from \((0,0)\) to \((1,1)\) and can be expressed as \((x, y) = (t, t)\) for \(t\) ranging from 0 to 1. This simplifies calculations because instead of varying both \(x\) and \(y\) independently, you just vary \(t\).

• The given \((x, y) = (1+2s, 1)\) describes the second segment where \(s\) ranges from 0 to 1.

Effective parametrization is an essential first step in solving many calculus problems, including line integrals.

Dot Product

The dot product is a way to multiply two vectors, which results in a scalar number rather than another vector. It is calculated by multiplying corresponding components of the vectors, and then summing those products. This is especially useful in line integrals to calculate how much one vector acts in the direction of another.

For example, in the step-by-step solution, \( \int_{0}^{1} (2t,0) \cdot (dt,dt) \) is calculated. Here, \(2t\) from the vector field \(\mathbf{G}\) and \(dt\) from the derivative \(d\mathbf{r}\) are multiplied together, and since the second component is zero, it simplifies to \(2t \, dt\).

• The formula for the dot product of vectors \(\mathbf{a} = (a_1, a_2)\) and \(\mathbf{b} = (b_1, b_2)\) is \(a_1b_1 + a_2b_2\).

Computing the dot product repeatedly in sections of a line integral helps determine the integral itself, showing how different parts of a vector field influence the path.

Calculus Solutions

Solving problems in calculus often involves a series of steps breaking down complex expressions into simpler calculations. In the problem presented, the calculus solutions involve calculating a line integral over a specified path.

Fundamentally, the goal is to find how a vector field behaves along a curve. This involves breaking down the path into segments, parametrizing them, and computing integrals for each segment. From the solution, we saw that the total line integral over \(C_2\) is \(1 + 6 = 7\). Here each calculation step aligns with accurately transitioning the variable bounds, simplifying dot products, and integrating.

• Aligning calculations with sections—parametrization, dot products, and integration steps—ensures correct calculus solutions.

Mastering such problems by understanding each segment distinctly can immensely help in grasping more advanced calculus concepts involving vector fields and line integrals.