This question is from my exercises in differential topology:
(i) $\alpha = yz dx+ xzdy + xy dz$
(ii) $\beta = xdx+ x^2 y^2 dy+ yz dz$
and (iii) $\gamma = 2 x y^2 dx \wedge dy+ z dy\wedge dz$.
Determine which of them is closed and exact.
Attempt: (a) I have done .it is both closed and exact as $\alpha= d(xyz)$.
(b) I am not able to make significant progress.how can I write $d \beta $ in terms of $dx\wedge dy$.
(c) I am not sure which results should I use to be able to manipulate $\wedge$'s so I am clueless on (c).
Can you please help me with (b) and (c)?
Calculating $d\alpha$ of some differential form $\alpha$ has a computational procedure, you need to just know some rules: if $$ \alpha = \sum_{1\leq i_1 < i_2 < \dots <i_k \leq n} f_{i_1,i_2,\dots,i_k}(x_1,x_2,\dots,x_n) dx_{i_1} \wedge dx_{i_2} \wedge \dots \wedge dx_{i_k} \tag{1} $$ is a differential $k$-form, then: $$ d\alpha = \sum_{1\leq i_1 < i_2 < \dots <i_k \leq n} \sum_{i=1}^n (\frac{\partial f_{i_1,i_2,\dots,i_k}}{\partial x_i} dx_i )\wedge dx_{i_1} \wedge dx_{i_2} \wedge \dots \wedge dx_{i_k} $$ you also should know that $dx_i\wedge dx_i = 0$. this means when you are calculating $d\alpha$ you don't need to calculate $\frac{\partial}{\partial x_i}$ for the terms that already has $dx_i$. and by notation in (1), we should reorder the differential terms to be in increasing order. now lets implement this on your $\beta$ : $$ \require{cancel} d \beta = \cancelto{0}{\frac{\partial (x)}{\partial y}}dy\wedge dx +\cancelto{0}{\frac{\partial (x)}{\partial z}}dz\wedge dx + \frac{\partial (x^2y^2)}{\partial x}dx\wedge dy +\cancelto{0}{\frac{\partial (x^2y^2)}{\partial z}} dz \wedge dy + \\ \cancelto{0}{\frac{\partial (zy)}{\partial x}} dx \wedge dz + \frac{\partial (zy)}{\partial y} dy\wedge dz \\ = (2xy^2)dx\wedge dy+zdy \wedge dz $$ since this is an exercise it's better to do $\gamma$ yourself with above rules. for example for first term you have $dx\wedge dy$ you should just take it's partial derivative with respect to $z$.
for checking the exactness you should solve some kind of system of PDEs. consider $\beta$ for instance since it's a 1-form we need to find a 0-form $\tau$ such that $d\tau =\beta$. a zero form is a smooth function ($C^{\infty}$) so it has a form of $\tau=f(x,y,z)$.($f$ is a smooth function) if $\beta$ is $d\tau$ we should have: $$ \frac{\partial f}{\partial x} = x, \quad \frac{\partial f}{\partial y} = x^2y^2, \quad \frac{\partial f}{\partial z} = yz $$ you can see that this system doesn't have a solution, [Edit 1] because for example start with solving first equation: $$ f(x,y,z) = \frac{1}{2}x^2 + g(y,z) $$ so if you plug this into second equation you'll get : $$ \frac{\partial g}{\partial y} =x^2y^2 \rightarrow g(y,z)=\frac{1}{3}x^2y^3+h(z) $$ which is a contradiction since $g$ can't be a function of $x$. therefore such $\tau$ doesn't exist and $\beta$ is not exact.
for $\gamma$ the system is a little more complicated since you should look for a 1-form. if $\gamma = d \tau$, and since $\tau$ is 1-form consider : $$ \tau = f_1(x,y,z)dx+f_2(x,y,z)dy+f_3(x,y,z)dz $$ it's a good exercise to calculate $d\tau$, but the final result is : $$ d\tau = (\frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y})dx\wedge dy + (\frac{\partial f_3}{\partial x} - \frac{\partial f_1}{\partial z})dx\wedge dz + (\frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z})dy\wedge dz \\ = 2xy^2dx\wedge dy + z dy\wedge dz $$ so you should solve vector-valued first-order system of PDEs: $$ \left\{ \begin{array}{l} \frac{\partial f_2}{\partial x} - \frac{\partial f_1}{\partial y} =2xy^2 \\ \frac{\partial f_3}{\partial x} - \frac{\partial f_1}{\partial z} =0 \\ \frac{\partial f_3}{\partial y} - \frac{\partial f_2}{\partial z} = z \end{array}\right. $$ it has some degree of freedom (3 equations, partial derivative of 3 functions) so you can put some extra assumption(s) and then try to solve it. for example consider $f_1=0$ then $f_2=x^2y^2$ and $f_3=yz$ would work.