微分形式の計算2

(1)
$$\begin{array}{rl}\mathrm{d}(\omega \wedge \xi )& =\mathrm{d}(x{y}^2\sin x\cos z\mathrm{d}x\wedge \mathrm{d}z+xy\mathrm{d}y\wedge \mathrm{d}z)\\ & =(-2xy\sin x\cos z+y)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z\end{array}$$

(2)
$$\begin{array}{rl}\mathrm{d}\mathrm{d}\omega & =\mathrm{d}(2y\sin x\cos z\mathrm{d}y\wedge \mathrm{d}x–y^2\sin x\sin z\mathrm{d}z\wedge \mathrm{d}x\\ & =-2y\sin x\sin z\mathrm{d}z\wedge \mathrm{d}y\wedge \mathrm{d}x–2y\sin x\sin z\mathrm{d}y\wedge \mathrm{d}z\wedge \mathrm{d}x\\ & =0\end{array}$$