Let's walk through this problem step-by-step. We are given a curve $\gamma$ parameterized in $\mathbb{R}^3$ with parametric equations:

$x = a \cos t, \quad y = a \sin t, \quad z = bt, \quad \text{for} \quad 0 \leq t \leq 2\pi.$

The task is to compute the line integral:

$\int_{\gamma} \mathbf{F} \cdot d\mathbf{r},$

where $\mathbf{F}(x, y, z) = \langle y - z, \, z - x, \, x - y \rangle$.

Step 1: Find $\mathbf{r}(t)$ and $\mathbf{r}'(t)$

The curve $\gamma$ is parameterized by:

$\mathbf{r}(t) = \langle a \cos t, \, a \sin t, \, bt \rangle.$

Now, we take the derivative with respect to $t$ to find $\mathbf{r}'(t)$:

$\mathbf{r}'(t) = \left\langle -a \sin t, \, a \cos t, \, b \right\rangle.$

Step 2: Express $\mathbf{F}(\mathbf{r}(t))$

Now, substitute the parametric equations $x = a \cos t$, $y = a \sin t$, and $z = bt$ into the components of $\mathbf{F}(x, y, z) = \langle y - z, \, z - x, \, x - y \rangle$.

• First component: $y - z = a \sin t - bt$,
• Second component: $z - x = bt - a \cos t$,
• Third component: $x - y = a \cos t - a \sin t$.

Thus,

$\mathbf{F}(\mathbf{r}(t)) = \langle a \sin t - bt, \, bt - a \cos t, \, a \cos t - a \sin t \rangle.$

Step 3: Compute the dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$

Now, we take the dot product of $\mathbf{F}(\mathbf{r}(t))$ with $\mathbf{r}'(t) = \langle -a \sin t, \, a \cos t, \, b \rangle$.

$\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = (a \sin t - bt)(-a \sin t) + (bt - a \cos t)(a \cos t) + (a \cos t - a \sin t)(b).$

Simplify the terms:

$= -a^2 \sin^2 t + abt \sin t + a^2 \cos^2 t - abt \cos t + ab \cos t - ab \sin t.$

Using the identity $\sin^2 t + \cos^2 t = 1$, the terms $-a^2 \sin^2 t + a^2 \cos^2 t$ simplify to $a^2$. Thus, we get:

$a^2 + abt(\sin t - \cos t) + ab(\cos t - \sin t).$

Notice that the terms $ab(\cos t - \sin t)$ cancel out with $abt(\sin t - \cos t)$ when integrated over the interval $t \in [0, 2\pi]$. Therefore, the dot product simplifies to:

$\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = a^2.$

Step 4: Evaluate the Line Integral

Now, we integrate the constant $a^2$ over the interval $t \in [0, 2\pi]$:

$\int_0^{2\pi} a^2 \, dt = a^2 \int_0^{2\pi} 1 \, dt = a^2 [t]_0^{2\pi} = a^2 (2\pi - 0) = 2\pi a^2.$

Final Answer

The value of the line integral is:

$\boxed{2\pi a^2}.$

Do you have any questions or need further details?

Here are 5 related questions for further exploration:

1. How would the integral change if $b$ were replaced by a function of $t$?
2. What if the vector field $\mathbf{F}$ were $\langle y + z, z + x, x + y \rangle$? How would the computation change?
3. Can Green’s theorem or Stokes’ theorem be applied to this problem in any way?
4. What is the geometric meaning of the integral for this parameterized curve?
5. How does the orientation of the curve affect the result?

Tip: When computing line integrals, always simplify the dot product as much as possible before integrating to save effort!