Given a surface bundle $S_g\to E \xrightarrow{f} S^1\vee S^1$, where $S_g$ is a closed surface of genus $g$. I want to calculate $H^3(E)$ by applying Mayer-Vietoris. Name the two circles of $S^1\vee S^1$ by $A$ and $B$ and define $U:=f^{-1}(A), V:=f^{-1}(A)$, which are $S_g$-bundles over $S^1$. Note $U\cap V\cong S_g$, thus by Mayer-Vietoris we have
$...\to H^2(U)\oplus H^2(V)\to H^2(S_g)=\mathbb Z\to H^3(E)\to H^3(U)\oplus H^3(V)=\mathbb Z^2\to H^3(U\cap V)\cong H^3(S_g)=1$
Now if $H^2(U)\oplus H^2(V)\to H^2(S_g)=\mathbb Z$ is trivial, then $H^3(E)=\mathbb Z^3$; if $H^2(U)\oplus H^2(V)\to H^2(S_g)=\mathbb Z$ is surjective, then $H^3(E)=\mathbb Z^2$.
I am not sure if the rank of $H^3(E)$ depends or it is always a constant?