Answer
Result: \[ \int (x^2 dx + x y dy + z^2 dz) = \frac{\pi^6}{192}. \]
Work Step by Step
Step 1: The path of integration is given by the parametrized curve: \[ \begin{align*} x &= \sin(t), \\ y &= \cos(t), \\ z &= t^2, \end{align*} \] for \(0 \leq t \leq \frac{\pi}{2}\). Therefore, the path integral: \[ \begin{align*} \int (x^2 dx + x y dy + z^2 dz) &= \int_{0}^{\frac{\pi}{2}} \sin^2(t)\cos(t)dt + \sin(t)\cos(t)(-\sin(t))dt + \frac{t^4}{2}2tdt \\ &= \int_{0}^{\frac{\pi}{2}} 2t^5\cos(t)dt \\ &= [2\frac{t^6}{6}]_0^{\frac{\pi}{2}} \\ &= \frac{1}{3}\left(\frac{\pi^6}{64} - 0\right) \\ &= \frac{\pi^6}{192}. \end{align*} \] Result: \[ \int (x^2 dx + x y dy + z^2 dz) = \frac{\pi^6}{192}. \]
