Math Problem Statement

explain

explain

Solution

This excerpt provides a mathematical definition and explanation of the line integral of a scalar function over a smooth curve in three-dimensional space. Let’s break it down:


Definition of the Line Integral

If f(x,y,z)f(x, y, z) is a scalar field (a function of position in 3D space), and the curve CC is given parametrically by: r(t)=g(t)i+h(t)j+k(t)k,atb,\mathbf{r}(t) = g(t)\mathbf{i} + h(t)\mathbf{j} + k(t)\mathbf{k}, \quad a \leq t \leq b, then the line integral of ff over CC is defined as: Cf(x,y,z)ds=limnk=1nf(xk,yk,zk)Δsk,\int_C f(x, y, z) \, ds = \lim_{n \to \infty} \sum_{k=1}^n f(x_k, y_k, z_k) \Delta s_k, where:

  • Δsk\Delta s_k is the arc length of a small segment of the curve CC between points.
  • This integral computes a weighted sum of ff over CC, where f(x,y,z)f(x, y, z) is evaluated at sample points on CC.

Condition for Smooth Curves

If the curve CC is smooth (meaning its velocity v(t)=drdt\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} is continuous and non-zero), and the function ff is continuous on CC, the integral exists.


Arc Length Substitution

To compute this integral, we use the arc length formula: s(t)=atv(τ)dτ,s(t) = \int_a^t |\mathbf{v}(\tau)| \, d\tau, where:

  • v(τ)=(dgdτ)2+(dhdτ)2+(dkdτ)2|\mathbf{v}(\tau)| = \sqrt{\left(\frac{dg}{d\tau}\right)^2 + \left(\frac{dh}{d\tau}\right)^2 + \left(\frac{dk}{d\tau}\right)^2}, the magnitude of the velocity vector.

Differentiating this with respect to tt gives ds=v(t)dtds = |\mathbf{v}(t)| \, dt.


Rewriting the Line Integral

Using the arc length substitution, the line integral becomes: Cf(x,y,z)ds=abf(g(t),h(t),k(t))v(t)dt.\int_C f(x, y, z) \, ds = \int_a^b f(g(t), h(t), k(t)) |\mathbf{v}(t)| \, dt. This is shown as Equation (2) in the image. It transforms the line integral into an ordinary definite integral over the parameter tt:

  • g(t),h(t),k(t)g(t), h(t), k(t) give the x,y,zx, y, z-coordinates of the curve CC.
  • v(t)|\mathbf{v}(t)| accounts for the arc length scaling.

Direction of Integration

  • The parameter tt defines a direction along the curve CC.
  • The integral is evaluated from t=at = a to t=bt = b, ensuring that the curve is traversed in the correct direction.

Conclusion

The line integral computes the total "weight" of ff along CC, weighted by the arc length. The parameterization ensures the computation works regardless of how the curve is defined, as long as it's smooth and properly oriented.


Questions to Explore Further

  1. Why is the velocity magnitude v(t)|\mathbf{v}(t)| necessary for transforming the line integral?
  2. How does the line integral differ if f(x,y,z)f(x, y, z) is a vector field instead of a scalar field?
  3. What happens if the curve CC is not smooth? How does it affect the line integral?
  4. Can you compute a line integral if the parameter tt is reversed (i.e., bab \to a)?
  5. How does the concept of line integrals relate to physical quantities like work or flux?

Tip

When working with line integrals, always carefully analyze the curve's parameterization and ensure v(t)|\mathbf{v}(t)| is computed correctly. Missteps here can lead to incorrect results.

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Math Problem Analysis

Mathematical Concepts

Line Integrals
Vector Calculus
Arc Length
Parametric Curves
Scalar Fields

Formulas

\int_C f(x, y, z) \, ds = \lim_{n \to \infty} \sum_{k=1}^n f(x_k, y_k, z_k) \Delta s_k
s(t) = \int_a^t |\mathbf{v}(\tau)| \, d\tau
\int_C f(x, y, z) \, ds = \int_a^b f(g(t), h(t), k(t)) |\mathbf{v}(t)| \, dt

Theorems

Line Integral Theorem
Fundamental Theorem of Calculus for Line Integrals

Suitable Grade Level

College-level (Undergraduate)