I'm having this problem and I have no idea how to figure it out.
In $\mathbb{R}^3$, we denote a linear operator $*$ : $Ω^1$($\mathbb{R}^3$) → $Ω^2$($\mathbb{R}^3$) defined by:
$∗dx = dy ∧ dz$ ,
$ ∗dy = dz ∧ dx,$
$∗dz = dx ∧ dy$
a) If $f$:$\mathbb{R}^3$→$\mathbb{R}$ is a smooth function, show that $d(∗df) = (∆f)dx ∧ dy ∧ dz$, where $∆f=∇ · ∇f$ is the Laplace operator of $f$.
b) Using Stokes Theorem, calculate $\int_s$ $\omega$
if $\omega$=$*df$
with $f=(x-y+2z)^2$ and $S$ an ellipsoid of equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ oriented in a way such that its orientation is compatible with its domain of $\mathbb{R}^3$
I would be grateful to anyone who has some ideas, thanks a lot!
Here are some hints and ideas. For both parts (a) and (b), you should use the formula $$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dx$$ For part (a): You should try to evaluate $d(\ast df)$. First find $df$, then $\ast df$, then $d(\ast df)$. Hopefully the result will be $\Delta f \,dx \wedge dy \wedge dz$.
For part (b): You should draw the given ellipsoid $S$, and ask yourself whether you can think of $S$ as the boundary surface of some solid $E$. Then Stokes' Theorem will give you $$\int_S \omega = \int_{\partial E} \ast df = \int_E d(\ast df).$$ Part (a) gives you a nice formula for $d(\ast df)$, which should help in evaluating the integral.