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The problem from the image reads:

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

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

### Step 1: Parameterize the Curve
The given curve is:
\[
\mathbf{r}(t) = t\mathbf{i} + t\mathbf{j} + t\mathbf{k}.
\]
This means that \( x = t \), \( y = t \), and \( z = t \), so the coordinates of the curve at any point are \( (t, t, t) \).

### Step 2: Express the Scalar Field Along the Curve
Substitute \( x = t \), \( y = t \), and \( z = t \) into the scalar field \( f(x, y, z) = \sqrt{3}(x^2 + y^2 - z^2) \):
\[
f(t, t, t) = \sqrt{3}(t^2 + t^2 - t^2) = \sqrt{3}(t^2).
\]
So along the curve, \( f(x, y, z) \) becomes \( f(t) = \sqrt{3} t^2 \).

### Step 3: Compute the Differential Arc Length \( ds \)
The differential arc length \( ds \) for the curve \( \mathbf{r}(t) \) is given by:
\[
ds = |\mathbf{r}'(t)| \, dt.
\]
First, compute the derivative of \( \mathbf{r}(t) \):
\[
\mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i} + t\mathbf{j} + t\mathbf{k}) = \mathbf{i} + \mathbf{j} + \mathbf{k}.
\]
The magnitude of \( \mathbf{r}'(t) \) is:
\[
|\mathbf{r}'(t)| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.
\]
Thus, \( ds = \sqrt{3} \, dt \).

### Step 4: Set Up the Line Integral
The line integral of the scalar field \( f(x, y, z) \) over the curve is:
\[
\int_C f(x, y, z) \, ds = \int_{1}^{2} f(t) |\mathbf{r}'(t)| \, dt.
\]
Substitute \( f(t) = \sqrt{3} t^2 \) and \( |\mathbf{r}'(t)| = \sqrt{3} \):
\[
\int_1^2 \sqrt{3} t^2 \cdot \sqrt{3} \, dt = 3 \int_1^2 t^2 \, dt.
\]

### Step 5: Solve the Integral
The integral of \( t^2 \) is:
\[
\int t^2 \, dt = \frac{t^3}{3}.
\]
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 \( 7 \).

Would you like any more details or clarifications?

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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 \( t \) 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.

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.

This page explains how to compute the line integral of the scalar field f(x, y, z) = √3(x^2 + y^2 - z^2) over the parametric curve r(t) = t*i + t*j + t*k, for t in [1, 2]. It covers key concepts in vector calculus, such as parameterization, differential arc length, and scalar fields.

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