Problem 10
Question
Evaluate \(\int_{C}(x-y+z-2) d s\) where \(C\) is the straight-line segment \(x=t, y=(1-t), z=1,\) from \((0,1,1)\) to \((1,0,1)\)
Step-by-Step Solution
Verified Answer
The value of the integral is \(-\sqrt{2}\).
1Step 1: Parameterize the Curve
The curve C is a line segment from (0,1,1) to (1,0,1). We can parameterize this line segment as follows:\[x = t, \quad y = 1 - t, \quad z = 1\]where \( t \) varies from 0 to 1. This parameterization represents the straight line from the point \((0,1,1)\) to \((1,0,1)\).
2Step 2: Express the Differential Element
The differential element \( ds \) for the line segment can be expressed as the magnitude of the derivative of the position vector with respect to the parameter \( t \):\[\frac{d\mathbf{r}}{dt} = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle = \langle 1, -1, 0 \rangle.\]Thus, the differential arc length is:\[ds = \sqrt{(1)^2 + (-1)^2 + (0)^2} \, dt = \sqrt{2} \, dt.\]
3Step 3: Substitute into the Integral
Substitute the parameterized variables and the expression for \( ds \) into the integral:\[\int_{0}^{1} (x - y + z - 2) \, ds = \int_{0}^{1} (t - (1-t) + 1 - 2) \, \sqrt{2} \, dt.\]Simplify the integrand:\[(t - 1 + t + 1 - 2) = 2t - 2.\]
4Step 4: Evaluate the Integral
Insert the simplification from Step 3 and evaluate:\[\int_{0}^{1} (2t - 2) \sqrt{2} \, dt = \sqrt{2} \int_{0}^{1} (2t - 2) \, dt.\]Compute the integral:\[= \sqrt{2} \left[ \left( t^2 - 2t \right) \right]_{0}^{1} = \sqrt{2} \left[ (1^2 - 2 \cdot 1) - (0^2 - 2 \cdot 0) \right] = \sqrt{2}(-1) = -\sqrt{2}.\]
5Step 5: Conclusion
The value of the line integral, after evaluating the expression, is \(-\sqrt{2}\). This concludes the computation of the integral over the straight-line segment C.
Key Concepts
Parameterized CurvesArc Length and its DifferentialUnderstanding Differential Elements
Parameterized Curves
When we need to evaluate line integrals, the first step is often to parameterize the curve. Parameterization means expressing a curve using a parameter, usually denoted by a variable like \( t \). In our exercise, the curve \( C \) is a straight line segment from \((0,1,1)\) to \((1,0,1)\).
To parameterize this line segment, we use the equations:
In general, parameterization is useful for different types of curves and is especially handy when dealing with complex shapes, as it translates curves into functions of a single variable. This makes the curve much easier to handle analytically, especially in calculus.
To parameterize this line segment, we use the equations:
- \( x = t \)
- \( y = 1 - t \)
- \( z = 1 \)
In general, parameterization is useful for different types of curves and is especially handy when dealing with complex shapes, as it translates curves into functions of a single variable. This makes the curve much easier to handle analytically, especially in calculus.
Arc Length and its Differential
The concept of arc length is pivotal when working with parameterized curves, especially in line integrals. When we have a curve parameterized by \( t \), the arc length element \( ds \) is the tiny segment of curve length corresponding to an infinitesimal change in \( t \).
To find \( ds \), we first derive the position vector \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) with respect to \( t \). For our line segment, it is:
The magnitude of this derivative vector gives us the differential arc length: \[ ds = \, \sqrt{(1)^2 + (-1)^2 + (0)^2} \, dt = \sqrt{2} \, dt. \]This formula tells us how much the length of the curve changes with a tiny increment in \( t \), which is crucial for computing line integrals since it represents the contribution to the integral per unit change in parameter.
To find \( ds \), we first derive the position vector \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) with respect to \( t \). For our line segment, it is:
- \( \frac{dx}{dt} = 1 \)
- \( \frac{dy}{dt} = -1 \)
- \( \frac{dz}{dt} = 0 \)
The magnitude of this derivative vector gives us the differential arc length: \[ ds = \, \sqrt{(1)^2 + (-1)^2 + (0)^2} \, dt = \sqrt{2} \, dt. \]This formula tells us how much the length of the curve changes with a tiny increment in \( t \), which is crucial for computing line integrals since it represents the contribution to the integral per unit change in parameter.
Understanding Differential Elements
In calculus, differential elements represent infinitesimally small quantities which, when integrated, provide macroscopic properties like area, volume, or in our case, line integrals.
For line integrals, the differential element typically refers to \( ds \), or the tiny length of the curve at a parameterized point. We compute it from derivatives of the parameterization, allowing us to transform the integral along the curve into a standard integral with respect to \( t \).
The derivative vector \( \frac{d\mathbf{r}}{dt} = \langle 1, -1, 0 \rangle \) represents components of the instantaneous rate of change for \( x \), \( y \), and \( z \) along the curve. Combining these components gives the magnitude, producing \( ds = \sqrt{2} \, dt \).
When we incorporate \( ds \) into our integral, it alters the integral to measure along the curve rather than just along the \( t \)-parameter. Without this differential element, we wouldn't accurately account for each segment's length contribution in the integral, especially important for curves that are not straight lines.
For line integrals, the differential element typically refers to \( ds \), or the tiny length of the curve at a parameterized point. We compute it from derivatives of the parameterization, allowing us to transform the integral along the curve into a standard integral with respect to \( t \).
The derivative vector \( \frac{d\mathbf{r}}{dt} = \langle 1, -1, 0 \rangle \) represents components of the instantaneous rate of change for \( x \), \( y \), and \( z \) along the curve. Combining these components gives the magnitude, producing \( ds = \sqrt{2} \, dt \).
When we incorporate \( ds \) into our integral, it alters the integral to measure along the curve rather than just along the \( t \)-parameter. Without this differential element, we wouldn't accurately account for each segment's length contribution in the integral, especially important for curves that are not straight lines.
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