I'm studying readings in geometry own my own and this question is confusing me.
Let $w$ be the 1-form on $\mathbb{R}^3$ defined by $$w=6 x y \cos(z) dx + 3 x^2 \cos(z) dy - 3 x^2 y \sin(z) dz$$ $a)$Show that $w$ is closed.
$b$)Find a function $f$ such that $w=df$
For the $a)$ part I need to show that the function is exact. But I can show only 2 variable $(x,y)$ equations. Do I need to differentiate $z$ as a both functions of $x$ and $y$ ?
For $b)$ sadly I have no idea.
Thanks!
Indeed, part $a)$ follows form part $b)$, as $d^2 = 0$.
However, for part $a)$, you have to prove that $dw = 0$. You can compute that using the fact that $d$ is linear and that $d(g dx) = dg \wedge dx$, for every smooth function $g$.
For instance, \begin{align}d(6xy\cos(z)dx) &= 6 \cdot \left( y\cos(z)dx + x\cos(z)dy - xy\sin(z)dz \right) \wedge dx \\ &= 6x\cos(z) dy \wedge dz - 6xy \sin(z) dz \wedge dx = \\ &= 6x\cos(z) dy \wedge dz + 6xy \sin(z) dx \wedge dz, \end{align} as $dx \wedge dx = 0$ and $dx \wedge dz = - dz \wedge dx$. You can similarly compute the remaining terms, and conclude that $dw = 0$.
For $b)$, you know that for every smooth function $f$, $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz, $$ so, imposing $df = w$, you get the following system of PDEs: $$\frac{\partial f}{\partial x} = 6xy\cos(z), \frac{\partial f}{\partial y} = 3x^2 \cos(z), \frac{\partial f}{\partial z} = -3x^2 y \sin(z).$$ You can integrate the first equation to get that $$f(x,y,z) = 3x^2y\cos(z) + T(y,z), $$ for some other smooth function $T$. Plugging this into the other equations, you will get that $T$ is constant, so you can choose $f(x,y,z) = 3x^2y\cos(z)$.