1. **State the problem:** We need to find the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = x^2 \mathbf{i} - yz \mathbf{j} + x \cos z \mathbf{k}$ and the path $C$ is given by the parametric equations $x = t^2$, $y = t$, $z = \pi t$. 2. **Recall the formula:** The line integral of a vector field $\mathbf{F}$ along a curve $C$ parameterized by $\mathbf{r}(t)$ from $t=a$ to $t=b$ is $$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt$$ 3. **Parameterize $\mathbf{r}(t)$ and compute $\mathbf{r}'(t)$:** $$\mathbf{r}(t) = \langle t^2, t, \pi t \rangle$$ $$\mathbf{r}'(t) = \langle 2t, 1, \pi \rangle$$ 4. **Substitute into $\mathbf{F}$:** $$x = t^2, \quad y = t, \quad z = \pi t$$ $$\mathbf{F}(\mathbf{r}(t)) = \langle (t^2)^2, -t \cdot \pi t, t^2 \cos(\pi t) \rangle = \langle t^4, -\pi t^2, t^2 \cos(\pi t) \rangle$$ 5. **Compute the dot product $\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$:** $$t^4 \cdot 2t + (-\pi t^2) \cdot 1 + t^2 \cos(\pi t) \cdot \pi = 2t^5 - \pi t^2 + \pi t^2 \cos(\pi t)$$ 6. **Determine the limits for $t$:** Since the problem does not specify, assume $t$ goes from 0 to 1. 7. **Set up the integral:** $$\int_0^1 \left(2t^5 - \pi t^2 + \pi t^2 \cos(\pi t)\right) dt$$ 8. **Integrate term by term:** - $\int_0^1 2t^5 dt = 2 \cdot \frac{t^6}{6} \Big|_0^1 = \frac{1}{3}$ - $\int_0^1 -\pi t^2 dt = -\pi \cdot \frac{t^3}{3} \Big|_0^1 = -\frac{\pi}{3}$ - $\int_0^1 \pi t^2 \cos(\pi t) dt$ requires integration by parts. 9. **Compute $I = \int_0^1 t^2 \cos(\pi t) dt$:** Let $u = t^2$, $dv = \cos(\pi t) dt$. Then $du = 2t dt$, $v = \frac{\sin(\pi t)}{\pi}$. $$I = u v \Big|_0^1 - \int_0^1 v du = \left. t^2 \frac{\sin(\pi t)}{\pi} \right|_0^1 - \int_0^1 \frac{\sin(\pi t)}{\pi} 2t dt = \frac{1}{\pi} \sin(\pi) - 0 - \frac{2}{\pi} \int_0^1 t \sin(\pi t) dt$$ Since $\sin(\pi) = 0$, the first term is zero. 10. **Compute $J = \int_0^1 t \sin(\pi t) dt$:** Let $u = t$, $dv = \sin(\pi t) dt$. Then $du = dt$, $v = -\frac{\cos(\pi t)}{\pi}$. $$J = u v \Big|_0^1 - \int_0^1 v du = \left. -\frac{t \cos(\pi t)}{\pi} \right|_0^1 + \int_0^1 \frac{\cos(\pi t)}{\pi} dt = -\frac{1 \cdot \cos(\pi)}{\pi} + 0 + \frac{1}{\pi} \int_0^1 \cos(\pi t) dt$$ 11. **Evaluate $\int_0^1 \cos(\pi t) dt$:** $$\int_0^1 \cos(\pi t) dt = \frac{\sin(\pi t)}{\pi} \Big|_0^1 = \frac{\sin(\pi)}{\pi} - 0 = 0$$ 12. **So, $J = -\frac{\cos(\pi)}{\pi} = -\frac{-1}{\pi} = \frac{1}{\pi}$** 13. **Back to $I$:** $$I = -\frac{2}{\pi} J = -\frac{2}{\pi} \cdot \frac{1}{\pi} = -\frac{2}{\pi^2}$$ 14. **Therefore,** $$\int_0^1 \pi t^2 \cos(\pi t) dt = \pi I = \pi \left(-\frac{2}{\pi^2}\right) = -\frac{2}{\pi}$$ 15. **Sum all parts:** $$\frac{1}{3} - \frac{\pi}{3} - \frac{2}{\pi}$$ **Final answer:** $$\int_C \mathbf{F} \cdot d\mathbf{r} = \frac{1}{3} - \frac{\pi}{3} - \frac{2}{\pi}$$