Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The value of the line integral is \( \frac{1}{3} \).

Step 1: Parametrize the Line Segment

To find a parametrization for the line segment from \((1,2,1)\) to \((2,1,0)\), we use the formula \( \mathbf{r}(t) = (1,2,1) + t((2,1,0)-(1,2,1)) \) where \( t \) ranges between \(0\) and \(1\). This gives \( \mathbf{r}(t) = (1+t, 2-t, 1-t) \).

Step 2: Determine Derivatives for the Parametrization

Calculate the derivatives \( \frac{d\mathbf{r}}{dt} = (1, -1, -1) \). These derivatives will be used when substituting into the line integral.

Step 3: Substitute into the Line Integral

Substitute \( x = 1+t \), \( y = 2-t \), and \( z = 1-t \) into the integrand \( x+y+z \) makes \((1+t)+(2-t)+(1-t) = 4-t\). Similarly, express the whole integral in terms of \( t \).

Step 4: Adjust the Integral Terms

The other integral terms become:- \( x \, dy \) translates to \( (1+t)(-1)dt = -(1+t)dt \)- \( -yz \, dz \) translates to \(-(-(2-t)(1-t)) = -(2-t)(1-t)dt \)

Step 5: Combine and Simplify the Integral Expression

Combine all parts to get the line integral:\[ \int_{0}^{1}[(4-t)\cdot 1 - (1+t) - (2-t)(1-t)]dt \].Simplify the expression:\[ = \int_{0}^{1} [4-t - 1 - t - 2(1-t) + t(1-t)] dt \] = \[ \int_{0}^{1} [1 - 2t + t^2] dt \].

Step 6: Evaluate the Integral

Calculate the integral:\[ \int_{0}^{1} [1 - 2t + t^2] dt = \int_{0}^{1} 1 \, dt - 2\int_{0}^{1} t \, dt + \int_{0}^{1} t^2 \, dt \].\[ = [t]_{0}^{1} - 2\left[\frac{t^2}{2}\right]_{0}^{1} + \left[\frac{t^3}{3}\right]_{0}^{1} \].Calculate each term: \[ =[1 - 0] - 2\left[\frac{1}{2} \right] + \left[\frac{1}{3} \right] \].Simplify to get: \[ 1 - 1 + \frac{1}{3} = \frac{1}{3} \].

Key Concepts

Parametrization of a Curve

Parametrization is an essential concept when dealing with line integrals because it helps express the points on a curve in a simple, continuous manner. When you have a line segment, like from point (1, 2, 1) to point (2, 1, 0), parametrizing involves creating a vector function that traces the path from the starting point to the end.

In our exercise, we used a linear parametrization: \[ \mathbf{r}(t) = (1 + t, 2 - t, 1 - t) \] where \( t \) is a parameter ranging from 0 to 1. This tells us how far along the line segment we are. When \( t = 0 \), we are at the starting point (1, 2, 1), and when \( t = 1 \), we are at the end point (2, 1, 0). By applying this approach, we express the entire line segment as a function of \( t \), allowing us to transform the problem into a manageable mathematical form.

Directional Derivatives

Directional derivatives are another integral part of understanding line integrals, particularly when functions change along a path. Though not explicitly required to solve the problem, the concept explains how a function changes as we move from point to point.

When calculating the line integral, understanding how the directional derivatives work is essential. The derivative of the parametrization, \( \frac{d\mathbf{r}}{dt} = (1, -1, -1) \), tells us the direction of movement along the line for an infinitesimally small change in \( t \). This is critical information because it influences how integrals are evaluated along the curve. Essentially, it dictates the 'direction' in which you're evaluating the derivative of the position, relative to each variable, over the line segment in the parameter space.

Integral Calculus

Integral calculus is the backbone of solving line integrals, like the one in the exercise. Line integrals extend the concept of integrals to integrating functions over a curve, rather than just along the x-axis.

In the step-by-step solution, integral calculus takes center stage when we substitute the parametrized values into the integral expression, converting a multi-variable expression into a one-variable integral. This is crucial because it simplifies the integration process:

• First, you're computing a definite integral for each term of the expression, adjusting the limits according to the parameter \( t \).
• Next, perform term-wise integration, allowing us to evaluate expressions like \( \int_{0}^{1}(1 - 2t + t^2) dt \).

Breaking down complex expressions into solvable segments is why integral calculus is powerful for these integrals. It shows that with systematic steps, any complicated function can be tackled.

Vector Calculus

Vector calculus extends regular calculus to multi-dimensional spaces, like the three-dimensional spaces discussed in line integrals. It deals not just with magnitudes, as in scalar calculus, but also with directions, giving it an additional layer of complexity and applicability.

In our example, understanding vector calculus helped to achieve a more thorough analysis of the problem. As each component of the vector function \( \mathbf{r}(t) \) influences its direction, line integrals incorporate vectors in several ways:

• Vectors denote directions and magnitudes in space, which are crucial when handling physical quantities that cannot be represented by mere numbers.
• Components \( (x, y, z) \) and derivatives \( \frac{d\mathbf{r}}{dt} \) give a comprehensive view of how the curve behaves throughout its domain.

Vector calculus includes these and many more concepts that help in not just integrating along lines, but also dealing with surfaces and volumes later on in higher mathematics.