Integrate The Function F = X 3y+ Z Over The Line Segment From The Point (0,0,0) To The Point (1,1,1).
When dealing with line integrals in multivariable calculus, understanding how to evaluate integrals of scalar functions along a specified curve is essential. One common problem involves integrating a function over a line segment connecting two points in three-dimensional space. In this article, we will explore how to evaluate the line integral of the scalar function \( F(x, y, z) = x^3 y + z \) along the line segment from the point \( (0, 0, 0) \) to \( (1, 1, 1) \). This step-by-step approach will deepen your understanding of parameterization, setting up the integral, and performing the calculations necessary to arrive at the solution.
---
Understanding the Problem
The goal is to compute the line integral:
\[
\int_{C} F(x, y, z) \, ds
\]
where \( C \) is the line segment from \( (0, 0, 0) \) to \( (1, 1, 1) \), and the scalar function is:
\[
F(x, y, z) = x^3 y + z
\]
This integral essentially sums the values of \( F \) along the path \( C \), weighted by the differential arc length \( ds \).
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Key Concepts in Line Integrals
Before delving into the calculation, let's review some fundamental concepts:
1. Parameterization of a Curve
To evaluate a line integral, we first parameterize the curve \( C \). For a line segment between two points, a simple parameterization uses a linear interpolation:
\[
\mathbf{r}(t) = \mathbf{r}0 + t(\mathbf{r}1 - \mathbf{r}_0), \quad t \in [0, 1]
\]
where:
- \( \mathbf{r}_0 = (0, 0, 0) \)
- \( \mathbf{r}_1 = (1, 1, 1) \)
Thus,
\[
\mathbf{r}(t) = (0 + t(1 - 0), 0 + t(1 - 0), 0 + t(1 - 0)) = (t, t, t)
\]
This parameterization is valid for \( t \in [0, 1] \).
2. Computing \( ds \)
The differential arc length \( ds \) can be expressed in terms of the parameter \( t \):
\[
ds = |\mathbf{r}'(t)| \, dt
\]
where
\[
\mathbf{r}'(t) = \frac{d\mathbf{r}}{dt}
\]
and
\[
|\mathbf{r}'(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}
\]
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Step-by-Step Solution
Let's now proceed step-by-step to evaluate the integral.
Step 1: Parameterize the Curve
As established:
\[
\mathbf{r}(t) = (t, t, t), \quad t \in [0, 1]
\]
which yields:
\[
x = t, \quad y = t, \quad z = t
\]
Step 2: Find \( \mathbf{r}'(t) \) and \( |\mathbf{r}'(t)| \)
Differentiating each component:
\[
\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 1, \quad \frac{dz}{dt} = 1
\]
Therefore,
\[
|\mathbf{r}'(t)| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
\]
Step 3: Rewrite the Scalar Function \( F \) in Terms of \( t \)
Substitute the parameterization into \( F \):
\[
F(\mathbf{r}(t)) = (t)^3 \times t + t = t^4 + t
\]
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Step 4: Set Up the Line Integral
The line integral becomes:
\[
\int{t=0}^{1} F(\mathbf{r}(t)) |\mathbf{r}'(t)| \, dt = \int0^1 (t^4 + t) \times \sqrt{3} \, dt
\]
which simplifies to:
\[
\sqrt{3} \int_0^1 (t^4 + t) \, dt
\]
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Step 5: Compute the Integral
Calculate the integral:
\[
\int0^1 t^4 \, dt = \left[ \frac{t^5}{5} \right]0^1 = \frac{1}{5}
\]
and
\[
\int0^1 t \, dt = \left[ \frac{t^2}{2} \right]0^1 = \frac{1}{2}
\]
Therefore,
\[
\sqrt{3} \left( \frac{1}{5} + \frac{1}{2} \right)
\]
Combine the fractions:
\[
\frac{1}{5} + \frac{1}{2} = \frac{2}{10} + \frac{5}{10} = \frac{7}{10}
\]
The final value of the line integral:
\[
\boxed{
\int_{C} F \, ds = \sqrt{3} \times \frac{7}{10} = \frac{7 \sqrt{3}}{10}
}
\]
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Summary and Final Result
By parameterizing the line segment from \( (0, 0, 0) \) to \( (1, 1, 1) \), expressing the function \( F \) in terms of the parameter, computing the differential arc length, and performing the integral, we've determined that:
\[
\boxed{
\int_{C} (x^3 y + z) \, ds = \frac{7 \sqrt{3}}{10}
}
\]
This process exemplifies how line integrals of scalar functions are evaluated in three-dimensional space, emphasizing the importance of parameterization, substitution, and integration techniques.
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Applications of Line Integrals in Physics and Engineering
Understanding how to compute line integrals such as this is fundamental in various scientific fields, including:
- Physics: Calculating work done by a force field along a path.
- Engineering: Determining flux and other properties in fluid dynamics and electromagnetism.
- Mathematics: Analyzing scalar and vector fields, studying properties like potential functions.
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Conclusion
Line integrals are powerful tools for analyzing functions along curves in space. The approach involves:
- Choosing an appropriate parameterization of the curve.
- Expressing the function in terms of the parameter.
- Calculating the differential arc length.
- Setting up and evaluating the integral.
Mastering these steps provides a solid foundation for tackling more complex problems in multivariable calculus and related disciplines.
