设 Gamma 是曲线 x = e^t cos t, y = e^t sin t, z = e^t 上相应于 t 从 0 变化到 1 的这段弧, 则曲线积分 int_(Gamma) (1)/(x^2 + y^2 + z^2) ds 等于: A. (sqrt(3))/(2) (e^-1 - 1)B. (sqrt(3))/(2) (1 - e^-2)C. (sqrt(3))/(2) (1 - e^-1)D. (3)/(2) (1 - e^-2)

为了求解曲线积分 $\int_{\Gamma} \frac{1}{x^2 + y^2 + z^2} \, ds$，其中 $\Gamma$ 是由参数方程 $x = e^t \cos t$, $y = e^t \sin t$, $z = e^t$ 给出的曲线，且 $t$ 从 0 变化到 1，我们需要按照以下步骤进行： 1. **计算 $x^2 + y^2 + z^2$:** \[ x^2 = (e^t \cos t)^2 = e^{2t} \cos^2 t \] \[ y^2 = (e^t \sin t)^2 = e^{2t} \sin^2 t \] \[ z^2 = (e^t)^2 = e^{2t} \] \[ x^2 + y^2 + z^2 = e^{2t} \cos^2 t + e^{2t} \sin^2 t + e^{2t} = e^{2t} (\cos^2 t + \sin^2 t + 1) = e^{2t} (1 + 1) = 3e^{2t} \] 2. **计算弧长元素 $ds$:** \[ ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \] \[ \frac{dx}{dt} = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \] \[ \frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \] \[ \frac{dz}{dt} = e^t \] \[ \left( \frac{dx}{dt} \right)^2 = \left( e^t (\cos t - \sin t) \right)^2 = e^{2t} (\cos t - \sin t)^2 = e^{2t} (\cos^2 t - 2 \cos t \sin t + \sin^2 t) = e^{2t} (1 - 2 \cos t \sin t) \] \[ \left( \frac{dy}{dt} \right)^2 = \left( e^t (\sin t + \cos t) \right)^2 = e^{2t} (\sin t + \cos t)^2 = e^{2t} (\sin^2 t + 2 \sin t \cos t + \cos^2 t) = e^{2t} (1 + 2 \sin t \cos t) \] \[ \left( \frac{dz}{dt} \right)^2 = (e^t)^2 = e^{2t} \] \[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 = e^{2t} (1 - 2 \cos t \sin t) + e^{2t} (1 + 2 \sin t \cos t) + e^{2t} = e^{2t} (1 - 2 \cos t \sin t + 1 + 2 \sin t \cos t + 1) = e^{2t} \cdot 3 = 3e^{2t} \] \[ ds = \sqrt{3e^{2t}} \, dt = \sqrt{3} e^t \, dt \] 3. **将 $x^2 + y^2 + z^2$ 和 $ds$ 代入曲线积分:** \[ \int_{\Gamma} \frac{1}{x^2 + y^2 + z^2} \, ds = \int_{0}^{1} \frac{1}{3e^{2t}} \cdot \sqrt{3} e^t \, dt = \int_{0}^{1} \frac{\sqrt{3}}{3e^{2t}} e^t \, dt = \int_{0}^{1} \frac{\sqrt{3}}{3e^t} \, dt = \frac{\sqrt{3}}{3} \int_{0}^{1} e^{-t} \, dt \] 4. **计算积分:** \[ \int_{0}^{1} e^{-t} \, dt = -e^{-t} \Big|_{0}^{1} = -e^{-1} - (-e^0) = -e^{-1} + 1 = 1 - e^{-1} \] \[ \frac{\sqrt{3}}{3} \int_{0}^{1} e^{-t} \, dt = \frac{\sqrt{3}}{3} (1 - e^{-1}) = \frac{\sqrt{3}}{2} (1 - e^{-1}) \] 因此，曲线积分 $\int_{\Gamma} \frac{1}{x^2 + y^2 + z^2} \, ds$ 的值是 $\frac{\sqrt{3}}{2} (1 - e^{-1})$。 正确答案是 $\boxed{C}$。