Problem 19 - Thomas Calculus

Question

In Exercises 19 and 20 , find the work done by \(F\) in moving a particle once counterclockwise around the given curve. $$\mathbf{F}=2 x y^{3} \mathbf{i}+4 x^{2} y^{2} \mathbf{j}$$ \(C :\) The boundary of the "triangular" region in the first quadrant enclosed by the \(x\) -axis, the line \(x=1,\) and the curve \(y=x^{3}\)

Step-by-Step Solution

Answer

The work done by \(\mathbf{F}\) is \(\frac{4}{3} + \frac{1}{4} + \frac{12}{11}\).

Step 1: Describe the Boundary

The curve described is in the first quadrant, enclosed by the x-axis (y=0), the line x=1, and the curve y=x^3. The parameterization of the path is counterclockwise, starting along the x-axis from (0,0) to (1,0), then up along x=1 to (1,1), and then back to (0,0) along the curve y=x^3.

Step 2: Setup the Line Integrals

We need to compute three line integrals for each segment of the path. Let's denote them as C1, C2, and C3. Since the force is a vector field, we use the line integral formula: \[ W = \oint_C \mathbf{F} \, \cdot \, d\mathbf{r} \],where \(d\mathbf{r} \) is the differential arc length vector.

Step 3: Integrate Over the Segment Along the x-axis (C1)

For path C1, along the x-axis, F is evaluated with y=0: \[ \mathbf{F}(x,0) = 0 \cdot \mathbf{i} + 0 \cdot \mathbf{j} = 0 \].The line integral becomes zero:\[ \int_{0}^{1} 0 \cdot dx = 0 \].

Step 4: Integrate Over the Segment Along x=1 (C2)

For the vertical line, x=1 from (1,0) to (1,1), the parameterization is: \(x=1, y=t\), where \(t\) goes from 0 to 1, and thus:\[ \mathbf{F}(1,t) = 2 \cdot 1 \cdot t^3 \cdot \mathbf{i} + 4 \cdot 1^2 \cdot t^2 \cdot \mathbf{j} = 2t^3 \mathbf{i} + 4t^2 \mathbf{j} \]. Calculate the line integral:\[ \int_{0}^{1} (2t^3\cdot 0 + 4t^2\cdot 1) \, dt = \int_{0}^{1} 4t^2 \, dt = \frac{4}{3}.\]

Step 5: Integrate Over the Curve y=x^3 (C3)

For the curve y=x^3 from (1,1) to (0,0), the parameterization is \(y=x^3\), so:\[ \mathbf{F}(x, x^3) = 2x(x^3)^3\mathbf{i} + 4x^2(x^3)^2\mathbf{j} = 2x^7\mathbf{i} + 4x^8\mathbf{j}.\]The differential vector is \(d\mathbf{r} = dx\mathbf{i} + 3x^2dx\mathbf{j}\).The integral becomes:\[ \int_{1}^{0} (2x^7 \cdot 1 + 4x^8 \cdot 3x^2) \, dx = \int_{1}^{0} (2x^7 + 12x^{10}) \, dx.\]Solving gives:\[ \left[ -\frac{1}{4}x^8 - \frac{12}{11}x^{11} \right]_1^0 = \frac{1}{4} + \frac{12}{11}.\]

Step 6: Add the Results

Add the integrals from each segment to find the total work done by \(\mathbf{F}\): \[ 0 + \frac{4}{3} + \left( \frac{1}{4} + \frac{12}{11} \right) = \frac{4}{3} + \frac{1}{4} + \frac{12}{11}.\]Combine fractions and simplify to get the total work.

Key Concepts

Vector Fields

In the realm of calculus, vector fields are fascinating mathematical constructs that assign a vector to every point in a subset of space. A vector field is often used to represent physical quantities that have both a magnitude and direction, such as force fields or magnetic fields.

In the given exercise, we consider a vector field represented by the vector function \( \mathbf{F} = 2xy^3 \mathbf{i} + 4x^2y^2 \mathbf{j} \). Here, the components \( \mathbf{i} \) and \( \mathbf{j} \) refer to the standard unit vectors in the directions of the x-axis and y-axis, respectively.

The primary purpose of a vector field in this context is to model the force exerted at each point along a path.

• The component \( 2xy^3 \) applies force in the direction of the x-axis.
• Meanwhile, \( 4x^2y^2 \) applies force in the direction of the y-axis.

These components vary according to the positions \( x \) and \( y \), and play a crucial role in determining the work done by the force field along a specified path.

Parameterization

Parameterization is a powerful tool in calculus that allows us to describe paths or curves using a parameter, often denoted as \( t \). This technique simplifies the evaluation of integrals over curves, especially in vector fields.

In this problem, parameterization is employed along each segment of the given triangular path. Essentially, each segment is broken down and expressed using a suitable parameter:

• For the segment along the x-axis from \((0,0)\) to \((1,0)\), the parameterization uses \( x \) as \( t \), while keeping \( y=0 \) constant.
• Along the segment from \((1,0)\) to \((1,1)\) following the line \( x=1 \), \( t \) takes the place of \( y \), with \( t \in [0,1] \).
• The most complex part is the curve \( y=x^3 \), parameterized by \( x \) as \( t \) from \(1\) to \(0\), as we move backwards.

Breaking down paths with parameterization, we can integrate along each segment to find out how much work is done by the vector field on each path piece.

Work Done by a Force Field

One of the fundamental applications of line integrals in vector calculus is determining the work done by a force field along a path. The work done is found by integrating the dot product of the force vector and the differential element of the path.

In this exercise, the work done by the field \( \mathbf{F} \) is calculated over a closed loop by evaluating three separate line integrals corresponding to each path segment.

• For the first segment, along the x-axis, the integral evaluates to zero because the force field does not exert any force (both components become zero).
• The second segment, along \( x=1 \), requires evaluating \( \int_{0}^{1} 4t^2 \, dt \), going to \( \frac{4}{3} \) units of work.
• Finally, for the curve \( y=x^3 \), the approach involves finding the work by integrating expressions involving higher powers of \( x \), giving \( \frac{1}{4} + \frac{12}{11} \).

Therefore, the total work done by the force field around the loop becomes the sum of these individual results.

These calculations showcase how the principles of vector calculus bridge the abstract fields of mathematical integration with real-world applications like evaluating work across complex paths.