I am given the following differential form $\eta:=dx\wedge dy\wedge dz$ and they ask me to compute the contraction $\omega=\iota_{\nu}\eta$, where $\nu=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}$ is the normal exterior unitari vector field.
My result turns out to be $\omega=x dy\wedge dz-ydx\wedge dz +zdx\wedge dy$ and I just wanted to know if my calculations where correct since I have no way to verify my result.
Method I:
Using the short cut [1]
$$\tag{1} \iota_\nu\,(dx\wedge dy\wedge dz)=dx(\nu)\,(dy\wedge dz)\color{red}-dy(\nu)\,(dx\wedge dz)+dz(\nu)\,(dx\wedge dy) $$ and \begin{align} &dx(\frac{\partial}{\partial x})=dy(\frac{\partial}{\partial y})=dz(\frac{\partial}{\partial y})=1\,,\\ &dx(\frac{\partial}{\partial y})=dx(\frac{\partial}{\partial z})=0\,, dy(\frac{\partial}{\partial x})=dy(\frac{\partial}{\partial z})=0\,, dz(\frac{\partial}{\partial x})=dz(\frac{\partial}{\partial y})=0\,,\tag{2} \end{align} gives you directly $$\tag{3} \iota_\nu\,(dx\wedge dy\wedge dz)=x\,dy\wedge dz\color{red}-y\,dx\wedge dz +z\,dx\wedge dy\,. $$ Method II:
Assuming that $\eta=dx\wedge dy\wedge dz$ is the volume form of $\mathbb R^3$ we have the Hodge star as $$\tag{4} \star\,dx=dy\wedge dz,\quad\star\,dy=dz\wedge dx,\quad\star\,dz=dx\wedge dy\,. $$ This allows to write (1) as \begin{align} \iota_\nu\,(dx\wedge dy\wedge dz)&=dx(\nu)\,(\star\,dx)\color{red}{+}dy(\nu)\,(\star\,dy)+dz(\nu)\,(\star\,dz)\\[2mm] &=\star\,(dx(\nu)\,dx+dy(\nu)\,dy+dz(\nu)\,dz)\,.\tag{5} \end{align} "Lowering indices" of an arbitrary vector field $\nu=\nu^x\frac{\partial}{\partial x}+\nu^y\frac{\partial}{\partial y}+\nu^z\frac{\partial}{\partial z}$ by the musical isomorphism gives the one-form $$\tag{6} \nu_\flat=\nu_x\,dx+\nu_y\,dy+\nu_z\,dz\,. $$ Observing that, in $\mathbb R^3\,,$ we have $\nu_x=\nu^x,\nu_y=\nu^y,\nu_z=\nu^z\,,$ and further that $$ dx(\nu)=\nu^x\,,\quad dy(\nu)=\nu^y\,,\quad dz(\nu)=\nu^z $$ holds we can write (5) very neatly as $$\tag{7}\boxed{\quad\phantom{\Big|} \iota_\nu(dx\wedge dy\wedge dz)=\star\,\nu_\flat\,.\quad} $$ Using this with $\nu=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}$ gives you $\nu_\flat=x\,dx+y\,dy+z\,dz$ which directly leads to (3).
Method III:
By definition: \begin{align}\tag{8} \omega:=\iota_\nu\,\eta(\boldsymbol{u},\boldsymbol{v}):=\eta(\nu,\boldsymbol{u},\boldsymbol{v}) \end{align} where $\boldsymbol{u},\boldsymbol{v}$ are arbitrary vector fields. To check that $\omega$ is of the form $$\tag{9} \omega=x\,dy\wedge dz-y\,dx\wedge dz +z\,dx\wedge dy $$ it might be a good exercise to
write out all wedge products \begin{align} dx\wedge dy&=dx\otimes dy-dy\otimes dx\,,\\ &\vdots\\ dx\wedge dy\wedge dz&=dx\otimes dy\otimes dz-dx\otimes dz\otimes dy\\ &~-dy\otimes dx\otimes dz+dy\otimes dz\otimes dx\\ &~+dz\otimes dx\otimes dy-dz\otimes dy\otimes dx\,; \end{align}
plug $\nu=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}$ into the first slot of $$\tag{10} \eta=dx\wedge dy\wedge dz\,, $$ that is \begin{align} \eta(\nu,\boldsymbol{u},\boldsymbol{v})&=x\,(dy\otimes dz)(\boldsymbol{u},\boldsymbol{v})-x\,(dz\otimes dy)(\boldsymbol{u},\boldsymbol{v})\\ &-y\,(dx\otimes dz)(\boldsymbol{u},\boldsymbol{v})+y\,(dz\otimes dx)(\boldsymbol{u},\boldsymbol{v})\\ &+ z\,(dx\otimes dy)(\boldsymbol{u},\boldsymbol{v}) - z\,(dy\otimes dx)(\boldsymbol{u},\boldsymbol{v})\,; \end{align} and
plug the basis vector fields $\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}$ in for $\boldsymbol{u}$ and $\boldsymbol{v}\,.$
This gives \begin{align} \omega(\frac{\partial}{\partial x},\frac{\partial}{\partial y})&=z\,, &\eta(\nu,\frac{\partial}{\partial x},\frac{\partial}{\partial y})&=z\,,\\ \omega(\frac{\partial}{\partial y},\frac{\partial}{\partial x})&=-z\,, &\eta(\nu,\frac{\partial}{\partial y},\frac{\partial}{\partial x})&=-z\,,\\ &\vdots \end{align} It is not hard to see that $\omega$ and $\eta(\nu,.,.)$ agree on all combinations. This proves again the equation (9).
The volume form $\eta$ and its contraction $\omega$ along $\nu$ have the interesting property that $$ d\omega=d(\iota_\nu\,\eta)=3\eta={\rm div}\nu\;\eta $$ which is easily seen from (9). The Stokes theorem $\int_{\partial V}\omega=\int_Vd\omega$ is here Gauss' divergence theorem and in this particular case it takes the form $$ \int_{\partial V}\iota_\nu\,\eta=\int_V{\rm div}\nu\;\eta\,. $$