Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{C}(x+y+z) d x+x d y-y z d z ; C\) is the line segment from \((1,2,1)\) to \((2,1,0)\).

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{13}{3}\).

Step 1: Parameterize the line segment

We can represent the line segment from the point \((1, 2, 1)\) to \((2, 1, 0)\) using a parameter \(t\), where \(0 \leq t \leq 1\). The parameterization is \(x(t) = 1+t\), \(y(t) = 2-t\), \(z(t) = 1-t\).

Step 2: Compute derivatives for parameterized paths

Calculate the derivatives of the parameterized expressions with respect to \(t\). We get: \(\frac{dx}{dt} = 1\), \(\frac{dy}{dt} = -1\), and \(\frac{dz}{dt} = -1\).

Step 3: Substitute into the integrand

Substitute the parameterized expressions for \(x, y, z\) and their derivatives into the given line integral: \[ \int_C (x+y+z) dx + x dy - yz dz = \int_{0}^{1} ((1+t) + (2-t) + (1-t))\cdot 1 + (1+t)(-1) - (2-t)(1-t)(-1) \ dt \]

Step 4: Simplify the integrand

Simplify the integrand: \[((1+t) + (2-t) + (1-t))\cdot 1 + (1+t)(-1) - (2-t)(1-t)(-1) = (4 - t) - (1+t) + (2-t)^2\] Expanding further, the integrand simplifies to: \[(4-t - 1-t + 4 - 4t + t^2) = (7 - 6t + t^2)\]

Step 5: Evaluate the integral

Integrate \(7 - 6t + t^2\) from \(0\) to \(1\): \[\int_{0}^{1} (7 - 6t + t^2) dt = \left[ 7t - 3t^2 + \frac{t^3}{3} \right]_{0}^{1} \] Evaluating from \(0\) to \(1\), we find: \[7(1) - 3(1)^2 + \frac{(1)^3}{3} - (0 - 0 + 0) \] \[= 7 - 3 + \frac{1}{3} = 4 + \frac{1}{3} = \frac{13}{3}\]

Step 6: Conclusion

The value of the line integral \(\int_C (x+y+z) dx + x dy - yz dz\) over the given path is \(\frac{13}{3}\).

Key Concepts

Parameterization

In order to evaluate a line integral over a path, we first need to describe the path as a function of a parameter, usually denoted as \( t \). This is known as parameterization. For the line segment from \((1, 2, 1)\) to \((2, 1, 0)\), we use \( t \) where \( 0 \leq t \leq 1 \). The parameterization involves expressing \( x, y, \) and \( z \) as functions of \( t \).

Here, we have:

• \( x(t) = 1 + t \)
• \( y(t) = 2 - t \)
• \( z(t) = 1 - t \)

This parameterization effectively describes every point along the line segment as \( t \) moves from 0 to 1, starting at \((1, 2, 1)\) and ending at \((2, 1, 0)\). Parameterization is crucial because it allows us to transform a multi-variable problem into a single-variable problem with respect to \( t \).

Path Derivatives

The next step in working with line integrals is to compute the derivatives of the parameterized path. These derivatives are needed to transform the integral into a form that's easier to evaluate. Here, we need to find the derivatives of the functions \( x(t) \), \( y(t) \), and \( z(t) \) with respect to \( t \).

The calculations give us:

• \( \frac{dx}{dt} = 1 \)
• \( \frac{dy}{dt} = -1 \)
• \( \frac{dz}{dt} = -1 \)

These derivatives represent the rate of change of each coordinate as \( t \) changes. They are vital for substituting into the integrand of the line integral, thereby enabling the conversion of the line integral into a definite integral over the interval \( [0, 1] \).

Integral Evaluation

With the parameterization and path derivatives established, these are substituted back into the line integral expression. The original expression is transformed into an integral involving the parameter \( t \). This integral will typically require simplification before it can be evaluated.

The integral expression, when substituted, results in:\[ \int_{0}^{1} ((1+t) + (2-t) + (1-t)) \cdot 1 + (1+t)(-1) - (2-t)(1-t)(-1) \, dt \]Simplifying the expression within the integral gives us:\[ 7 - 6t + t^2 \]This simplification is crucial as it prepares the integral for straightforward evaluation. Simplification is one of the potential hurdles in line integrals, but once achieved, the final integral can be handled using standard integration techniques.

Calculus

Finally, we use fundamental calculus to evaluate the simplified integral. With the expression \( 7 - 6t + t^2 \) ready, the integration process involves applying the rules of definite integrals over the interval \( [0, 1] \).

The integration process involves:

• \( 7t - 3t^2 + \frac{t^3}{3} \) evaluated from 0 to 1
• Plugging in the limits results in: \( 7 - 3 + \frac{1}{3} = \frac{13}{3} \)

Calculus provides the tools to take the final simplified expression to its final evaluated answer, giving the total accumulation of values along the path described. The computation shows how calculus allows us to find exact values for line integrals, demonstrating the power and utility of integral calculus in real-world applications.