Evaluating a surface integral Denote by $m$ the surface measure on the unit sphere $B$ in $\mathbb{C}^2$. Consider an unitary $2\times 2$ matrix $U$ and suppose we have a function $f\colon B\to \mathbb{C}$ which is $m$-integrable. Is it true that
$\int_B f(x)\,\mbox{d}x = \int_B f(Ux)\,\mbox{d}x?$
 A: Unitary matrices $U$ have the property that $<x,y>=<Ux,Uy>$, meaning they preserve the inner-product , i.e., the metric . Using a Real version of the Jacobian 
(since |Det(U)|=1) , may also give you an argument to this effect.
A: Yes the equation is true. As other people have pointed out, for $U\in U(2)$ the map $\mathbb{C}^2\ni x\mapsto Ux\in\mathbb{C}^2$ is an isometry of $\mathbb{C}^2$ w.r.t. the standard metric so it defines a map $T\colon S^3\to S^3$ by restriction, moreover the uniform measure on $S^3$ is preserved by $T$ (in order to view it, you can pick a definition of such measure on $S^3$ which uses the standard metric of $\mathbb{C}^2$ and then observe that this metric is preserved under a unitary transformation, hence also the measure is). Another possibility, already suggested, is to get your hands dirty with the change of variables formula and use the fact that the Jacobian of $T\colon S^3\to S^3$ is $1$ because $|\det U|=1$.
Notice that your equation generalizes to any dimension $n\geq 1$, i.e. for any $U\in U(n)$ and $f\in S^{2n-1}\to \mathbb{C}$. 
(Given that I contributed more confusion than anything, I suggest not to upvote this answer)
