【答案】 $$\text { （1）因为 }\left\{\begin{array}{l} x \mathrm{~d} x+y \mathrm{~d} y+z \mathrm{~d} z=0, \\ \mathrm{~d} y+\mathrm{d} z=0, \end{array} \text { 所以 } \mathrm{d} y=-\mathrm{d} z, x \mathrm{~d} x=-2 z \mathrm{~d} z \text {. 又因为在 } \Gamma \text { 上 } z \geqslant\right.$$
0 , 故 $z=\frac{\sqrt{2}}{2} \sqrt{1-x^2}$, 得 $x \mathrm{~d} x=-\sqrt{2\left(1-x^2\right)}$. 由此 $\pm\left\{-\sqrt{2\left(1-x^2\right)},-x, x\right\}$ 为 $\Gamma$ 上任 意一点 $(x, y, z)$ 处的切向量, 进而得与 $\Gamma$ 同向的切向量为 $\left\{\sqrt{2\left(1-x^2\right)}, x,-x\right\}$, 故切向 量的方向余弦为
$$\cos \alpha=\frac{\sqrt{2\left(1-x^2\right)}}{\sqrt{\left(\sqrt{\left.2\left(1-x^2\right)\right)^2+(x)^2+(-x)^2}\right.}}=\sqrt{1-x^2} \cdot \cos \beta=\frac{x}{\sqrt{2}}, \cos \gamma=\frac{-x}{\sqrt{2}} .$$

\begin{aligned} \int_{\Gamma} P \mathrm{~d} x+Q \mathrm{~d} y+R \mathrm{~d} z & =\int_{\Gamma}(P \cos \alpha+Q \cos \beta+R \cos \gamma) \mathrm{d} s \\ & =\int_{\Gamma}\left(\sqrt{1-x^2} P+\frac{x}{\sqrt{2}} Q-\frac{x}{\sqrt{2}} R\right) \mathrm{d} s \end{aligned}

(2) 由 (1) 以及 $z=\frac{1}{\sqrt{2}} \sqrt{1-x^2}$ 和 $y=-\frac{1}{\sqrt{2}} \sqrt{1-x^2}$ 得
\begin{aligned} I & =\int_{\Gamma} z \mathrm{~d} x+(x+\cos x) \mathrm{d} y+\mathrm{e}^{y^2} \mathrm{~d} z \\ & =\int_{\Gamma}\left[\sqrt{1-x^2} z+\frac{x}{\sqrt{2}}(x+\cos x)-\frac{x}{\sqrt{2}} \mathrm{e}^{y^2}\right] \mathrm{d} s \\ & =\int_{\Gamma}\left[\frac{1}{\sqrt{2}}\left(1-x^2\right)+\frac{1}{\sqrt{2}}\left(x^2+x \cos x\right)-\frac{x}{\sqrt{2}} \mathrm{e}^{\frac{1}{2}\left(1-x^2\right)}\right] \mathrm{d} s \\ & =\frac{1}{\sqrt{2}} \int_{\Gamma}\left[1+x \cos x-x \mathrm{e}^{\frac{1}{2}\left(1-x^2\right)}\right] \mathrm{d} s . \end{aligned}

$\Gamma$ 关于 $y() z$ 坐标面对称, 且 $x \cos x-x \mathrm{e}^{\frac{1}{2}\left(1-x^2\right)}$ 关于 $x$ 为奇函数, 故由曲线积分的奇偶对称 性可得 $\int_r\left[x \cos x-x \mathrm{e}^{\frac{1}{2}\left(1-x^2\right)}\right] \mathrm{d} s=0$. 所以
$$I=\frac{1}{\sqrt{2}} \int_{\Gamma} \mathrm{d} s=\frac{1}{\sqrt{2}} \cdot \Gamma \text { 的弧长 }=\frac{1}{\sqrt{2}} \pi \text {. }$$