To evaluate the line integral

$\int_C \nabla \phi \cdot d\mathbf{r}$

for the given scalar function $\phi(x, y, z) = \frac{x^2 + y^2 + z^2}{2}$ along the curve $C: \mathbf{r}(t) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle$ over the interval $t \in \left[ \frac{3\pi}{4}, \frac{5\pi}{4} \right]$, we can use both (a) a parametric approach, and (b) the Fundamental Theorem for line integrals.

Step (a): Direct Evaluation Using Parametric Description

1. Compute $\nabla \phi$:

   $\phi(x, y, z) = \frac{x^2 + y^2 + z^2}{2}$

   The gradient $\nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right)$ is:

   $\frac{\partial \phi}{\partial x} = x, \quad \frac{\partial \phi}{\partial y} = y, \quad \frac{\partial \phi}{\partial z} = z$

   Thus,

   $\nabla \phi = \langle x, y, z \rangle$

2. Parametrize $\nabla \phi$ along $C$:

   Since $\mathbf{r}(t) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle$, we substitute $x = \cos t$, $y = \sin t$, and $z = \frac{t}{\pi}$ into $\nabla \phi$:

   $\nabla \phi(\mathbf{r}(t)) = \langle \cos t, \sin t, \frac{t}{\pi} \rangle$

3. Compute $d\mathbf{r}$:

   Differentiate $\mathbf{r}(t)$ with respect to $t$:

   $\frac{d\mathbf{r}}{dt} = \langle -\sin t, \cos t, \frac{1}{\pi} \rangle$

   So, $d\mathbf{r} = \langle -\sin t, \cos t, \frac{1}{\pi} \rangle \, dt$.

4. Compute $\nabla \phi \cdot d\mathbf{r}$:

   Now, calculate the dot product $\nabla \phi(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}$:

   $\nabla \phi \cdot \frac{d\mathbf{r}}{dt} = \langle \cos t, \sin t, \frac{t}{\pi} \rangle \cdot \langle -\sin t, \cos t, \frac{1}{\pi} \rangle$

   Expanding this dot product:

   $= (\cos t)(-\sin t) + (\sin t)(\cos t) + \frac{t}{\pi} \cdot \frac{1}{\pi}$

   Simplify each term:

   $= -\cos t \sin t + \cos t \sin t + \frac{t}{\pi^2}$

   The terms $-\cos t \sin t$ and $\cos t \sin t$ cancel each other out, leaving:

   $= \frac{t}{\pi^2}$

5. Integrate over $t$:

   Now, integrate from $t = \frac{3\pi}{4}$ to $t = \frac{5\pi}{4}$:

   $\int_C \nabla \phi \cdot d\mathbf{r} = \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} \frac{t}{\pi^2} \, dt$

   This simplifies to:

   $= \frac{1}{\pi^2} \int_{\frac{3\pi}{4}}^{\frac{5\pi}{4}} t \, dt$

   Evaluating the integral:

   $= \frac{1}{\pi^2} \cdot \left[ \frac{t^2}{2} \right]_{\frac{3\pi}{4}}^{\frac{5\pi}{4}}$

   Substituting the limits:

   $= \frac{1}{\pi^2} \cdot \left( \frac{(5\pi/4)^2}{2} - \frac{(3\pi/4)^2}{2} \right)$

   $= \frac{1}{\pi^2} \cdot \frac{1}{2} \left( \frac{25\pi^2}{16} - \frac{9\pi^2}{16} \right)$

   Simplify further:

   $= \frac{1}{\pi^2} \cdot \frac{1}{2} \cdot \frac{16\pi^2}{16} \left( 25 - 9 \right)$