Problem 15 - Thomas Calculus

Question

Integrate \(f(x, y, z)=x+\sqrt{y}-z^{2}\) over the path from \((0,0,0)\) to \((1,1,1)(\) see accompanying figure) given by $$ \begin{array}{ll}{C_{1} :} & {\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}, \quad 0 \leq t \leq 1} \\ {C_{2} :} & {\mathbf{r}(t)=\mathbf{i}+\mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1}\end{array} $$ The paths of integration for Exercises 15 and 16

Step-by-Step Solution

Answer

The integral over the entire path is \(\frac{8}{3}\).

Step 1: Break the Integral into Two Segments

The integration is performed over two paths: \(C_1\) and \(C_2\). Each segment of the path requires a separate line integral.

Step 2: Parameterize the Path C_1

For the path \(C_1\), given \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j}\), the parameterization is \((x = t, y = t^2, z = 0)\) for \(0 \leq t \leq 1\).

Step 3: Compute the Integral over C_1

Calculate the line integral of \(f(x, y, z) = x + \sqrt{y} - z^2\) over \(C_1\). Substitute \(x=t\), \(y=t^2\), and \(z=0\) into \(f\): \[\int_{C_1} f(x, y, z) \, ds = \int_0^1 [t + \sqrt{t^2}] \, dt = \int_0^1 [t + t] \, dt.\]The above integral simplifies to:\[\int_0^1 2t \, dt = \left[ t^2 \right]_0^1 = 1.\]

Step 4: Parameterize the Path C_2

For the path \(C_2\), given \(\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t \mathbf{k}\), the parameterization is \((x = 1, y = 1, z = t)\) for \(0 \leq t \leq 1\).

Step 5: Compute the Integral over C_2

Calculate the line integral of \(f(x, y, z) = x + \sqrt{y} - z^2\) over \(C_2\). Substitute \(x=1\), \(y=1\), and \(z=t\) into \(f\): \[\int_{C_2} f(x, y, z) \, ds = \int_0^1 [1 + 1 - t^2] \, dt = \int_0^1 [2 - t^2] \, dt.\]The above integral simplifies to:\[\left[ 2t - \frac{t^3}{3} \right]_0^1 = \left(2 - \frac{1}{3}\right) = \frac{5}{3}.\]

Step 6: Combine Results from Both Paths

Add the results from both segments of the path:\[\int_{C_1} f(x, y, z) \, ds + \int_{C_2} f(x, y, z) \, ds = 1 + \frac{5}{3} = \frac{8}{3}.\]

Key Concepts

Path Parameterization

Path parameterization is a fundamental concept in line integrals, allowing us to express a curve using a single parameter, typically denoted as \( t \). This method simplifies the computation of integrals over paths, as it reduces multi-variable functions into functions of one variable.
Path parameterization requires us to describe each point along the curve as a function of \( t \). For instance, the path \( C_1 \) in the exercise is described by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \), translating to \( (x = t, y = t^2, z = 0) \).
With parameterization, different segments of a curve can be handled separately. Each segment might have a different parameterization, as seen with \( C_2 \), where points are parameterized by \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t \mathbf{k} \) or \( (x = 1, y = 1, z = t) \). This simplification is crucial as it transforms our multi-variable function \( f(x, y, z) \) into a more manageable single-variate function.

Integral Computation

Once paths are parameterized, the next step is integral computation. This involves calculating the line integral along each path segment. The primary aim is to integrate the function \( f(x, y, z) = x + \sqrt{y} - z^2 \) over the given paths.

• For \( C_1 \): Substitute \( x = t \), \( y = t^2 \), and \( z = 0 \) into \( f \), resulting in \( \int_{C_1} (t + t) \, dt = \int_0^1 2t \, dt \). After simplifying, the integral evaluates to \( 1 \).
• For \( C_2 \): Substitute \( x = 1 \), \( y = 1 \), and \( z = t \) into \( f \), which results in \( \int_{C_2} (2 - t^2) \, dt \). This simplifies further to \( \frac{5}{3} \) after evaluation.

The sum of the integrals from both paths gives the total integral of the function over the combined path. This step-by-step approach is a powerful technique to manage complexity in vector calculus.

Vector Calculus

The study of vector calculus is crucial for understanding line integrals, as it provides the tools needed to manipulate and evaluate functions along paths in three-dimensional space.
Line integrals in vector calculus often involve vectors described by \( \mathbf{r}(t) \). These vectors trace the path through space. In our path parameterizations, vectors \( \mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} \) for \( C_1 \) and \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k} \) for \( C_2 \) highlight this aspect.
Understanding vector fields and how they interact with paths is essential, as these fields determine the behavior of our function \( f(x, y, z) \) along its path. While the function in this exercise isn't a vector field, the process follows similar steps, showcasing the broader applications of line integrals in vector calculus.