To solve the given line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$, we proceed as follows:

Problem Breakdown:

1. Vector field: $\mathbf{F}(x, y, z) = x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}$
2. Parameterization of the curve $C$: $\mathbf{r}(t) = 2\sin(t)\mathbf{i} + 2\cos(t)\mathbf{j} + \frac{1}{2}t^2\mathbf{k}, \, 0 \leq t \leq \pi$
3. Differential displacement vector: $d\mathbf{r} = \frac{d\mathbf{r}}{dt}dt$

Steps to Evaluate:

1. Compute $d\mathbf{r}/dt$ from $\mathbf{r}(t)$: $\mathbf{r}(t) = (2\sin(t), 2\cos(t), \frac{1}{2}t^2)$ Differentiating: $\frac{d\mathbf{r}}{dt} = (2\cos(t), -2\sin(t), t)$ Hence, $d\mathbf{r} = (2\cos(t)\mathbf{i} - 2\sin(t)\mathbf{j} + t\mathbf{k})dt$.

2. Substitute $\mathbf{r}(t)$ into $\mathbf{F}(x, y, z)$: Using $\mathbf{r}(t) = (x, y, z) = (2\sin(t), 2\cos(t), \frac{1}{2}t^2)$, $\mathbf{F}(x, y, z) = (2\sin(t))^2\mathbf{i} + (2\cos(t))^2\mathbf{j} + \left(\frac{1}{2}t^2\right)^2\mathbf{k}$ Simplify: $\mathbf{F}(x, y, z) = 4\sin^2(t)\mathbf{i} + 4\cos^2(t)\mathbf{j} + \frac{1}{4}t^4\mathbf{k}.$

3. Compute the dot product $\mathbf{F} \cdot d\mathbf{r}$: Substitute $\mathbf{F}(x, y, z)$ and $d\mathbf{r}$ into the dot product: $\mathbf{F} \cdot d\mathbf{r} = \left(4\sin^2(t)\right)(2\cos(t)) + \left(4\cos^2(t)\right)(-2\sin(t)) + \left(\frac{1}{4}t^4\right)(t)$ Simplify each term: $\mathbf{F} \cdot d\mathbf{r} = 8\sin^2(t)\cos(t) - 8\cos^2(t)\sin(t) + \frac{1}{4}t^5.$ Factorize the trigonometric terms: $\mathbf{F} \cdot d\mathbf{r} = 8\sin(t)\cos(t)(\sin(t) - \cos(t)) + \frac{1}{4}t^5.$

4. Integrate over $t \in [0, \pi]$: $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^\pi \left[8\sin(t)\cos(t)(\sin(t) - \cos(t)) + \frac{1}{4}t^5\right] dt.$ Separate the integral: $\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^\pi 8\sin(t)\cos(t)(\sin(t) - \cos(t)) dt + \int_0^\pi \frac{1}{4}t^5 dt.$

5. Evaluate the trigonometric integral: Let $u = \sin(t)$, $du = \cos(t)dt$. Change limits accordingly ($u(0) = 0, u(\pi) = 0$): The term $\int_0^\pi 8\sin(t)\cos(t)(\sin(t) - \cos(t)) dt$ simplifies and evaluates to zero because the limits are symmetric.

6. Evaluate the polynomial integral: $\int_0^\pi \frac{1}{4}t^5 dt = \frac{1}{4} \cdot \frac{t^6}{6}\Big|_0^\pi = \frac{1}{24}\pi^6.$

Final Result:

$\int_C \mathbf{F} \cdot d\mathbf{r} = \frac{\pi^6}{24}.$

Would you like further details or explanations?
Here are 5 related questions to explore:

1. How is the parameterization of a curve used in evaluating line integrals?
2. What are the typical applications of line integrals in physics or engineering?
3. How do symmetry properties simplify integral computations?
4. How can we interpret $\mathbf{F} \cdot d\mathbf{r}$ geometrically?
5. What are the differences between line integrals of scalar and vector fields?

Tip: Always check for symmetry in trigonometric terms; it can greatly simplify integral evaluations!