Math Problem Statement

Find the line integral of f(x, y, z) = √3(x^2 + y^2 - z^2) over the curve r(t) = t*i + t*j + t*k, 1 <= t <= 2.

Solution

The problem from the image reads:

"Find the line integral of
f(x,y,z)=3(x2+y2z2)f(x, y, z) = \sqrt{3}(x^2 + y^2 - z^2) over the curve
r(t)=ti+tj+tk,1t2.\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k}, \quad 1 \leq t \leq 2.**

This is a line integral of a scalar field f(x,y,z)f(x, y, z) over a parametrized curve r(t)\mathbf{r}(t). Let's break it down step-by-step:

Step 1: Parameterize the Curve

The given curve is: r(t)=ti+tj+tk.\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k}. This means that x=tx = t, y=ty = t, and z=tz = t, so the coordinates of the curve at any point are (t,t,t)(t, t, t).

Step 2: Express the Scalar Field Along the Curve

Substitute x=tx = t, y=ty = t, and z=tz = t into the scalar field f(x,y,z)=3(x2+y2z2)f(x, y, z) = \sqrt{3}(x^2 + y^2 - z^2): f(t,t,t)=3(t2+t2t2)=3(t2).f(t, t, t) = \sqrt{3}(t^2 + t^2 - t^2) = \sqrt{3}(t^2). So along the curve, f(x,y,z)f(x, y, z) becomes f(t)=3t2f(t) = \sqrt{3} t^2.

Step 3: Compute the Differential Arc Length dsds

The differential arc length dsds for the curve r(t)\mathbf{r}(t) is given by: ds=r(t)dt.ds = |\mathbf{r}'(t)| \, dt. First, compute the derivative of r(t)\mathbf{r}(t): r(t)=ddt(ti+tj+tk)=i+j+k.\mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i} + t\mathbf{j} + t\mathbf{k}) = \mathbf{i} + \mathbf{j} + \mathbf{k}. The magnitude of r(t)\mathbf{r}'(t) is: r(t)=12+12+12=3.|\mathbf{r}'(t)| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}. Thus, ds=3dtds = \sqrt{3} \, dt.

Step 4: Set Up the Line Integral

The line integral of the scalar field f(x,y,z)f(x, y, z) over the curve is: Cf(x,y,z)ds=12f(t)r(t)dt.\int_C f(x, y, z) \, ds = \int_{1}^{2} f(t) |\mathbf{r}'(t)| \, dt. Substitute f(t)=3t2f(t) = \sqrt{3} t^2 and r(t)=3|\mathbf{r}'(t)| = \sqrt{3}: 123t23dt=312t2dt.\int_1^2 \sqrt{3} t^2 \cdot \sqrt{3} \, dt = 3 \int_1^2 t^2 \, dt.

Step 5: Solve the Integral

The integral of t2t^2 is: t2dt=t33.\int t^2 \, dt = \frac{t^3}{3}. Now evaluate the integral from 1 to 2: [ 3 \left[ \frac{t^3}{3} \right]_1^2 = \left[ t^3 \right]_1^2 = 2^3 - 1^3 = 8 - 1 = 7. ]

Final Answer:

The value of the line integral is 77.

Would you like any more details or clarifications?


Here are five related questions to explore further:

  1. What is the physical meaning of a line integral in the context of a scalar field?
  2. How would the solution change if the curve was different?
  3. What are the conditions for evaluating line integrals in vector fields versus scalar fields?
  4. How does the parameterization of a curve affect the evaluation of a line integral?
  5. How would this integral change if the limits of tt were from 0 to 1 instead?

Tip: For vector fields, line integrals require dot products with the curve's tangent vector, while for scalar fields, it's a simple multiplication of the field with the differential arc length.

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Math Problem Analysis

Mathematical Concepts

Line Integral
Scalar Field
Parametric Curve

Formulas

f(x, y, z) = √3(x^2 + y^2 - z^2)
r(t) = t*i + t*j + t*k
ds = |r'(t)| dt

Theorems

Line Integral of a Scalar Field

Suitable Grade Level

Undergraduate