Existence of diffeomorphism so that $\int (\phi^* f) \omega = 0$ Can someone help me with this question? 
Is a qual exam question and I have no idea how to tackle it.
Prove or disprove:
Let $f\in \mathcal{C}^\infty(S^n)$ be a smooth function and $x_1, x_2\in S^n$
be two points such that $f(x_1)<0<f(x_2)$.
Let $\omega$ be a smooth nonvanishing $n$-form. Then there is a diffeomorphism $\phi\colon S^n \rightarrow S^n$ such that $\int_{S^n} (\phi^* f)\omega =0$.
 A: Based on the hint:
Since $S^n$ is path connected, there is path $\gamma:[0,1]\to S^n$ such that $\gamma(0)=x_1,\gamma(1)=x_2$. We can take this path to be injective (think about $S^2$ using angular coordinates). Push the vector field $\dfrac{\partial}{\partial t}$ forward using $\gamma$ to a vector field $V$ on $\gamma([0,1])$. Since this is an embedded submanifold which is closed in $S^n$, we can extend this to a compactly supported vector field (also called $V$) defined on $S^n$ (see Lee, chapter 8).
Because $S^n$ is compact, $V$ admits a complete flow, i.e. a smooth 1-parameter family of diffeomorphisms $\Phi(t,\cdot)=\phi_t:S^n\to S^n$. Consider the map $$F=f\circ \Phi(\cdot,x_1):\mathbb{R}\to\mathbb{R}$$Then $F(0)=f(\Phi(0,x_1))=f(x_1)<0$ and $F(1)=f(\Phi(1,x_1))=f(x_2)>0$ (check that $\Phi(1,x_1)=x_2$). By the intermediate value theorem, there is a $t_0\in(0,1)$ with $F(t_0)=0$. Now consider the form $(\phi_{t_0}^{\ast}f)\omega$....
Still working on it.
A: The short answer was given by studiosus in the comment already.
If $\int f \omega = 0$, set $\phi  =\text{id}$ and we are done. Assume that $\int f \omega <0$ (If not, consider $-f$). Since $\omega$ is nonvanishing, We also assume that $\omega$ is positive, so
$$\mu(A) := \int_A \omega >0$$
for all nonempty open sets $A$. 
Let $x_0 \in \mathbb S^n$ such that $f$ is positive. Then by continuity there is an open ball $U$ centered at $x_0$ so that $f \ge c>0$ on $U$. Let $\{\phi_r\}_{r >0}$ be the one parameter family of diffeomorphisms so that $\phi_1 = \text{id}$, $\phi_s U \subset \phi_r U$ if $s<r$ and 
$$\bigcup_r \phi^{-1}_r U = \mathbb S^n\setminus \{-x_0\}.$$
(see the construction below) Note that 
$$g(r)= \int_{\mathbb S^n} (\phi_r^* f)\omega$$
is continuous and $g(1) <0$. Since $f$ is uniformly bounded below by $-m$, we have 
$$\begin{split}
\int_{\mathbb S^n} (\phi_r^* f)\omega &= \int_{\phi_r^{-1}\ U} (\phi_r^*f) \omega + \int_{\mathbb S^n\ \setminus \phi^{-1}_r\ U} (\phi_r^*f)\omega\\
&\ge c \mu(\phi^{-1}_r U) - m \mu( \mathbb S^n \setminus \phi^{-1}_r U) \\
&\to c \mu(\mathbb S^n) >0
\end{split}$$
as $r\to 0$, where $\mu(A) = \int_A \omega$. Thus there is $r_0$ so that 
$$\int_{\mathbb S^n} (\phi_{r_0}^* f)\omega =0.$$
To construct the family of diffeomorphisms, consider the stereographic projection at $\psi : \mathbb S^n \setminus \{-x_0\} \to \mathbb R^n$ at $x_0$. Let $U = \psi^{-1}(B_1)$, where $B_1$ is the unit ball in $\mathbb R^n$. Then define $\phi_r : \mathbb S^n \to \mathbb S^n$ by 
$$ \phi_r(x) =\begin{cases} \psi^{-1} (r\psi(x)) & \text{if }x\neq -x_0 \\ x_0 &\text{if }x = -x_0 .\end{cases}$$
Note that $\phi_r$ is smooth even at $-x_0$, since if you use the stereographic projection $\overline \psi$ at $-x_0$, you get (Since $\overline \psi \circ \psi (x) =\frac{x}{|x|^2}$)
$$\phi_r(x) = \begin{cases} \overline\psi^{-1} (\frac 1r \overline \psi(x)) &\text{if } x\neq  x_0 \\
x_0 & \text{if }x = x_0.\end{cases}$$
Note that you can also construct a one parameter family of diffeomorphisms using the gradient vector (with the standard metric on $\mathbb S^n$) of the function 
$$ F(x) = x\cdot x_0,$$
where the dot product is the standard one on $\mathbb R^{n+1}$. 
