1. **State the problem:** Calculate the line integral $$\int_C yz\,dx - xz\,dy + xy\,dz$$ along the curve $$C$$ defined by $$x=2\sin t$$, $$y=2\cos t$$, $$z=2t$$ for $$0 \leq t \leq \frac{\pi}{4}$$. 2. **Parameterize the integral:** Given $$x(t)=2\sin t$$, $$y(t)=2\cos t$$, $$z(t)=2t$$, compute derivatives: $$\frac{dx}{dt} = 2\cos t$$, $$\frac{dy}{dt} = -2\sin t$$, $$\frac{dz}{dt} = 2$$. 3. **Substitute into the integral:** The line integral becomes $$\int_0^{\pi/4} \left[y(t)z(t) \frac{dx}{dt} - x(t)z(t) \frac{dy}{dt} + x(t)y(t) \frac{dz}{dt}\right] dt$$. 4. **Calculate each term:** - $$y(t)z(t) \frac{dx}{dt} = (2\cos t)(2t)(2\cos t) = 8t \cos^2 t$$ - $$-x(t)z(t) \frac{dy}{dt} = -(2\sin t)(2t)(-2\sin t) = 8t \sin^2 t$$ - $$x(t)y(t) \frac{dz}{dt} = (2\sin t)(2\cos t)(2) = 8 \sin t \cos t$$ 5. **Sum the terms inside the integral:** $$8t \cos^2 t + 8t \sin^2 t + 8 \sin t \cos t = 8t(\cos^2 t + \sin^2 t) + 8 \sin t \cos t = 8t + 8 \sin t \cos t$$ 6. **Integral becomes:** $$\int_0^{\pi/4} (8t + 8 \sin t \cos t) dt = 8 \int_0^{\pi/4} t dt + 8 \int_0^{\pi/4} \sin t \cos t dt$$ 7. **Evaluate each integral:** - $$\int_0^{\pi/4} t dt = \left[\frac{t^2}{2}\right]_0^{\pi/4} = \frac{\pi^2}{32}$$ - Use identity $$\sin t \cos t = \frac{1}{2} \sin 2t$$, so $$\int_0^{\pi/4} \sin t \cos t dt = \frac{1}{2} \int_0^{\pi/4} \sin 2t dt = \frac{1}{2} \left[-\frac{\cos 2t}{2}\right]_0^{\pi/4} = -\frac{1}{4} (\cos \frac{\pi}{2} - \cos 0) = -\frac{1}{4} (0 - 1) = \frac{1}{4}$$ 8. **Combine results:** $$8 \times \frac{\pi^2}{32} + 8 \times \frac{1}{4} = \frac{8\pi^2}{32} + 2 = \frac{\pi^2}{4} + 2$$ **Final answer:** $$\boxed{\frac{\pi^2}{4} + 2}$$