Problem 24 - Thomas Calculus

Question

Evaluate $$\int_{C} x^{2} d x+y z d y+\left(y^{2} / 2\right) d z$$ along the line segment \(C\) joining \((0,0,0)\) to \((0,3,4).\)

Step-by-Step Solution

Answer

The line integral evaluates to 18.

Step 1: Parameterize the Line Segment

To evaluate the line integral, we need to parameterize the line segment from the point \((0,0,0)\) to \((0,3,4)\). We can use a parameter \(t\) such that \(0 \leq t \leq 1\). The parameterization is given by \(\mathbf{r}(t) = (0, 3t, 4t)\).

Step 2: Differentiate the Parameterization

Next, we need to differentiate the parameterized functions with respect to \(t\).\- \(\frac{dx}{dt} = 0\), since \(x = 0\).\- \(\frac{dy}{dt} = 3\), since \(y = 3t\).\- \(\frac{dz}{dt} = 4\), since \(z = 4t\).

Step 3: Substitute into the Line Integral

Now substitute the parameterization and its derivatives into the line integral: \[\int_{C} (x^{2} \, dx + yz \, dy + \frac{y^{2}}{2} \, dz) = \int_{0}^{1} (0^{2} \cdot 0 + (3t)(4t) \cdot 3 + \left(\frac{(3t)^{2}}{2}\right) \cdot 4) \, dt.\] Simplifying gives \[\int_{0}^{1} (12t^{2} \cdot 3 + \frac{9t^{2}}{2} \cdot 4) \, dt.\]

Step 4: Simplify the Integrand

Calculating inside the integral, we have: \(36t^{2}\) for \(12t^{2} \cdot 3\), and \(18t^{2}\) for \(\frac{9t^{2}}{2} \cdot 4\). Thus, the integral becomes \[\int_{0}^{1} (36t^{2} + 18t^{2}) \, dt = \int_{0}^{1} 54t^{2} \, dt.\]

Step 5: Evaluate the Integral

The integrand is a simple polynomial. Calculate the integral: \(\int_{0}^{1} 54t^{2} \, dt = 54 \int_{0}^{1} t^{2} \, dt = 54 \left[ \frac{t^{3}}{3} \right]_{0}^{1} = 54 \cdot \frac{1}{3} = 18.\)

Step 6: Conclude

The value of the line integral along the line segment from \((0, 0, 0)\) to \((0, 3, 4)\) is 18.

Key Concepts

Parameterization

Parameterization is the process of expressing a line or curve using a parameter, usually denoted as \( t \). In this context, we are parameterizing a line segment from the point \((0,0,0)\) to \((0,3,4)\). The parameter \( t \) ranges from 0 to 1.
You can think of parameterization as a way to write down the path of a journey in terms of time or some other simple variable.

• Start Point: \((0,0,0)\)
• End Point: \((0,3,4)\)
• Parameterization: \( \mathbf{r}(t) = (0, 3t, 4t) \)

By setting the parameter \( t \) to vary between 0 and 1, you smoothly transition from the start point to the end point. Each value of \( t \) gives a specific point on the line. This makes it easier to substitute values directly into integrals, especially for line integrals like in this problem.

Integral Calculus

Integral calculus is a branch of calculus focused on the concept of integration, which is essentially the process of finding areas, volumes, and summations under a curve. In a line integral context, we integrate over curves or paths in space.
Line integrals involve summing function values along a path, considering both the function value and the measure of the path. The given problem involves integrating a vector field along a parameterized path.

• The general form of line integral: \( \int_{C} f(x, y, z) \, ds \)
• In this exercise: \( \int_{C} x^{2} dx + yz dy + (y^{2}/2) dz \)

Each segment of the integral corresponds to one part of the vector field and how it interacts with the respective path differential \( dx \), \( dy \), or \( dz \). Integrating these parts while considering the parameterization of the path gives us the total integral value.

Polynomial Integration

Polynomial integration is a fundamental concept of calculus that deals with finding the integral of polynomial expressions. In this exercise, after parameterizing the line and substituting, we end up with a polynomial to integrate.In the line integral, advanced into \[ \int_{0}^{1} 54t^{2} \, dt \]we notice that the integrand is already a polynomial: \( 54t^2 \). The rules for integrating polynomials are straightforward.

• The integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} \) plus a constant.
• For our polynomial, integrate: \( 54 \int_{0}^{1} t^{2} dt \).
• Solve: \( 54 \left[ \frac{t^3}{3} \right]_{0}^{1} = 54 \cdot \frac{1}{3} \).

The final result after calculation is 18. This demonstrates the simplicity of polynomial integration, allowing us to quickly solve line integral problems involving polynomials.