Problem 15 - Thomas Calculus

Question

The paths of integration for Exercises 15 and 16 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 $$ \begin{array} { l l } { C _ { 1 } : } & { \mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } , \quad 0 \leq t \leq 1 } \\ { C _ { 2 } : } & { \mathbf { r } ( t ) = \mathbf { i } + \mathbf { j } + t \mathbf { k } , \quad 0 \leq t \leq 1 } \end{array} $$

Step-by-Step Solution

Answer

The total integral value is \(\frac{8}{3}\).

Step 1: Identify path segments

The problem specifies two path segments, \(C_1\) from \((0,0,0)\) to \((1,1,0)\), and \(C_2\) from \((1,1,0)\) to \((1,1,1)\). The integration needs to be performed along these paths separately and then summed up.

Step 2: Parameterize the paths

The paths are given as parametric equations. For \(C_1\), \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j},\; 0 \leq t \leq 1\). For \(C_2\), \(\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k},\; 0 \leq t \leq 1\). Interpret these paths to find the expressions for \(x(t), y(t), z(t)\).

Step 3: Determine differential elements

Calculate the differentials \(dx, dy, dz\). For \(C_1\), \(dx = dt\), \(dy = 2t dt\), \(dz = 0\). For \(C_2\), \(dx = 0\), \(dy = 0\), \(dz = dt\).

Step 4: Integrate over \(C_1\)

Substitute \(x(t), y(t), z(t)\) into the integrand. For \(C_1\), the function becomes \(f(x,y,z) = t + \sqrt{t^2} - 0^2 = 2t\). The integral is \(\int_0^1 2t dt = [t^2]_0^1 = 1\).

Step 5: Integrate over \(C_2\)

Similarly for \(C_2\), substitute back into the integrand. The function is \(f(x,y,z) = 1 + 1 - t^2 = 2 - t^2\). Integrate: \(\int_0^1 (2 - t^2) dt\). Compute this integral: \([2t - \frac{t^3}{3}]_0^1 = 2 - \frac{1}{3} = \frac{5}{3}\).

Step 6: Sum the integrals

Add the results of the integrals from steps 4 and 5: \(1 + \frac{5}{3} = \frac{8}{3}\).

Key Concepts

Parametric Equations

In the world of mathematical functions, parametric equations offer a unique way to describe a curve in space by using a parameter, typically denoted as \(t\). This parameter acts like an invisible hand that traces the path along which the vector lies.

For instance, consider the curve \(C_1\) in the exercise, which is represented by \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}\) for \(0 \leq t \leq 1\). Here, \(t\) guides how \(x\) and \(y\) vary as the curve is formed from the starting point \((0,0,0)\) to \((1,1,0)\). Similarly, \(C_2\) is described by \(\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k}\), moving from \((1,1,0)\) to \((1,1,1)\).

Parametric equations are immensely helpful because they allow us to express and calculate paths that are complex or impossible to describe using a single coordinate function. By expressing paths as parametric equations, we are able to perform calculus operations such as integration and differentiation, which are essential in many fields including physics, engineering, and computer graphics.

Differential Elements

Understanding differential elements is crucial when working with line integrals, as they are the building blocks that allow us to perform integration on a curve. A differential element represents a tiny piece of the curve produced by the parametric equations.

In this exercise, for path \(C_1\), we have the parametric forms \(x(t) = t\) and \(y(t) = t^2\), leading to differentials of \(dx = dt\) and \(dy = 2t dt\), respectively. For \(z\), since it remains unchanged along \(C_1\), we have \(dz = 0\). These differentials tell us how tiny changes in \(t\) translate to changes in \(x\), \(y\), and \(z\).

Similarly, for path \(C_2\), \(dx = 0\) and \(dy = 0\) since \(x\) and \(y\) are constants, and \(dz = dt\) because \(z\) changes with \(t\). Understanding these differential elements allows us to properly set up and evaluate integrals over these paths, as each element contributes to the total integral based on its path and direction.

Integration of Vector Fields

Line integrals allow us to integrate over vector fields, which can be visualized as assigning a vector quantity to every point along a path. This concept is especially useful in physics when dealing with quantities like force fields or fluid flow.

In our exercise, the function \(f(x, y, z) = x + \sqrt{y} - z^2\) is integrated over the paths \(C_1\) and \(C_2\). For \(C_1\), substituting the parametric equations given, \(f(x, y, z)\) simplifies to \(2t\). The integral of this over \(t\) from \(0\) to \(1\) is \([t^2]_0^1 = 1\).

For \(C_2\), the function reduces to \(2 - t^2\) upon substitution into the parametric form. Integrating this from \(0\) to \(1\) yields \(\left[2t - \frac{t^3}{3}\right]_0^1 = \frac{5}{3}\). The final result is the sum of these integrals, \(1 + \frac{5}{3} = \frac{8}{3}\).

Through such integrals, we capture the cumulative effect of the function over the path, which could represent physical work done by a force field or the total flow of a fluid, showing the importance of line integrals in real-world applications.