Let $f: B_1(0) \subset \mathbb{R}^4 \rightarrow \mathbb{R}$ such that $f(v) = f(x,y,z,w) = \tilde{f}(|(x,y)|, |(z,w)|)$, for some $\tilde{f}$. I have to prove there's a positive constant $C$ such that $$ \int_{B} f(v) dv = C \int_{D} \tilde{f}(s,t) s\, t\, ds dt, $$ where $D = \{(s,t) \in \mathbb{R}^2 :, s,t \geq 0, |(s,t)| < 1\}$.
My attempt: Consider $T : Q := [0,1] \times [0,1] \times [0,2\pi] \times [0,2\pi] \rightarrow \mathbb{R}^4$ definied by $$T(s,t, \theta, \varphi) = (s \cos \theta, s \sin \theta, t cos\varphi, t \sin \varphi).$$ So $B = T(Q)$. One can easily prove that $|J T| = st$. Thus $$ \int_B f(v) dv = \int_0^1 \int_0^1 \int_0^{2\pi} \int_0^{2\pi} \tilde{f}(|(s \cos \theta, s \sin \theta)|, |(t cos\varphi, t \sin \varphi)|) s\, t\, d \theta \varphi ds dt. $$ In other words, $$ \int_B f(v) dv = (2\pi)^2 \int_0^1 \int_0^1 \tilde{f} (s, t) s \, t \, d s dt. $$ My question is how to see that $\int_0^1\int_0^1 \tilde{f} (s, t) s \, t \, d s dt$ equals, up to a constant, to $\int_{D} \tilde{f}(s,t) s\, t\, ds dt$.