91Ó°ÊÓ

The given curve is parameterized by the given equations: \(x = 2\sin t, y = t, z = -2\cos t\). The interval for \(t\) is \(0 \leq t \leq \pi\).

The position vector \(\mathbf{r}(t)\) in terms of \(t\) is \(\mathbf{r}(t) = \langle 2\sin t, t, -2\cos t \rangle\). The derivative \(\mathbf{r}'(t) = \langle 2\cos t, 1, 2\sin t \rangle\).

The differential arc length \(ds\) is calculated as \(ds = \left\| \mathbf{r}'(t) \right\| dt\). This gives us \(ds = \sqrt{(2\cos t)^2 + 1^2 + (2\sin t)^2} dt = \sqrt{4\cos^2 t + 1 + 4\sin^2 t} dt = \sqrt{5} dt\).

Substitute \(x = 2\sin t\), \(y = t\), \(z = -2\cos t\), and \(ds = \sqrt{5} dt\) into the integral: \[ \int_{C} x y z ds = \int_{0}^{\pi} (2\sin t)(t)(-2\cos t) \sqrt{5} dt \] Simplify and evaluate:\[ \int_{0}^{\pi} -4\sqrt{5} t \sin t \cos t dt = -4\sqrt{5} \int_{0}^{\pi} t \sin t \cos t dt \]Using the identity \(\sin t \cos t = \frac{1}{2} \sin(2t)\), rewrite the expression: \[ -2\sqrt{5} \int_{0}^{\pi} t \sin(2t) dt \]Apply integration by parts. Let \(u = t\), \(dv = \sin(2t) dt\) which gives \(du = dt\), \(v = -\frac{1}{2}\cos(2t)\). Integrate by parts:\[ -2\sqrt{5} \left( \left. -\frac{t}{2}\cos(2t) \right|_{0}^{\pi} + \frac{1}{2} \int_{0}^{\pi} \cos(2t) dt \right) \] Compute the evaluated terms:\[ \left. -\frac{t}{2}\cos(2t) \right|_{0}^{\pi} = -\frac{\pi}{2}\cos(2\pi) + \frac{0}{2}\cos(0) = -\frac{\pi}{2} \]\[ \frac{1}{2}\int_{0}^{\pi} \cos(2t) dt = \frac{1}{4}\left(\sin(2t)\right)_{0}^{\pi} = 0 \]The integral evaluates to: \[-2\sqrt{5} \left( -\frac{\pi}{2} + 0 \right) = \pi\sqrt{5}\]

The value of the line integral is \(\pi\sqrt{5}\).