Simplification or properties of surface integral $\oint|\vec{U}(\vec{r})\times\hat{r}|^2$ over surface of sphere? Can we simplify the following surface integral in some way, or say anything about the properties it must have? 
$$
\oint_{\partial\mathbb{B}\left(\vec{0},R\right)} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \;ds\left(\vec{r}\right)
$$
$\vec{U}$ is a continuous vector field the value of which depends on the position $\vec{r} \in \mathbb{R}^3$, while $\mathbb{B}\left(\vec{0},R\right)$ is a sphere centered on the origin ($\vec{0}$) with radius $R$. The unit vector $\hat{r}$ is the unit normal (pointing outward) to the sphere at position $\vec{r}$ (so $\vec{r}=R\;\hat{r}$).
Is it, specifically, possible to say that this integral will be proportional to $R^2$, for instance?
Thanks...
Update: It has been pointed out that I need to provide some more details about the field $\vec{U}$, so here it is. Actually, the $\vec{U}$ I'm interested in is time dependent, and is defined as:
$$
\vec{U}\left(\vec{r}, t\right) = \iiint_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t-\frac{\left|\vec{r}-\vec{a}\right|}{c} \right)\;dV\left(\vec{a}\right)
$$
Here, $\mathbb{B}\left(\vec{0},R_s\right)$ is another sphere centered on the origin, with a radius $R_s < R$, while $c$ is a constant. $\vec{J}$ is another vector field which is not necessarily continuous, and has the following properties:
$$
\forall t, \quad \forall \vec{a} \in \partial\mathbb{B}\left(\vec{0},R_s\right)\qquad \vec{J}\left( \vec{a}, t \right) = 0
$$ 
$$
\forall t, \quad \forall \vec{a} \notin \mathbb{B}\left(\vec{0},R_s\right)\qquad \vec{J}\left( \vec{a}, t \right) = 0
$$
That is, $\vec{J}$ is zero on the surface of and outside $\mathbb{B}\left(\vec{0},R_s\right)$. 
Update 2: The following may or may not be true for my particular $\vec{U}$:
$$
\iiint_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t\right)\;dV\left(\vec{a}\right) = 0
$$
[Due to a brain freeze on my part, I had previously thought that this must be the case, but it isn't really. Sorry for the confusion there...] 
(This stuff comes from classical electrodynamics, as you can probably guess!)
 A: For simplicity of discussion, let us assume your $\vec{J}$ has only one relevant length scale $R_s$, i.e.
$$\left|\frac{\partial \vec{J}(\vec{a},t)}{\partial \vec{a}}\right| \sim~ O\left(\frac{|\vec{J}(\vec{a},t)|}{R_s}\right)$$
If that is the case, then $\vec{J}$ has only two relevant time scales $T_s$ and $T_c$.
$T_s = \frac{R_s}{c}$ is the time scale for a light beam to travel across the inner sphere of radius $R_s$. $T_c$ is the time scale for temporal changes of $\vec{J}$. i.e.
$$\left|\frac{\partial \vec{J}(\vec{a},t)}{\partial t}\right| \sim~ O\left(\frac{|\vec{J}(\vec{a},t)|}{T_s}\right)$$
If $T_c \gg T_s$, $\vec{J}$ is slowly varying in time and in the defining integral of $\vec{U}$, the $t$ dependence effectively goes away. In this limit,
$$
\vec{U}\left(\vec{r}, t\right) 
= \int_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t-\frac{\left|\vec{r}-\vec{a}\right|}{c} \right)\;dV\left(\vec{a}\right)
\rightarrow \int_{\mathbb{B}\left(\vec{0},R_s\right)} \vec{J}\left(\vec{a},t-\frac{\left|\vec{r}\right|}{c} \right)\;dV\left(\vec{a}\right)
= 0
$$
If $T_c \ll T_s$, then for any two points inside $\mathbb{B}(0,R_s)$ separated by a distance $c T_c$, the spatial variation of $\vec{J}$ is significantly smaller than the temporal variation. In the defining integral of $U$, nothing stop these two points from  destructively interfere with each other. One should expect $\vec{U}$ vanishes in this limit too.
This leaves us the case $T_c \sim O(T_s)$. Let's imagine we perform some sort of multipole expansion of the angular dependence of $\vec{J}$. Once again, the assumption of single length scale $R_s$ tells us in the multipole expansion, the lower term is more important.
If the multipole expansion is indeed dominated by the lowest term or $T_c$ is big (but not too big) compared to $T_s$, then for a point $\vec{r}$ at a distance $|\vec{r}| = R \gg R_s$, the leading contribution to $\vec{U}(\vec{r},t)$ should take the form:
$$
\begin{align}
\vec{U}\left(\vec{r}, t\right) \sim &
\int_{\mathbb{B}\left(\vec{0},R_s\right)} \frac{\partial\vec{J}}{\partial t}\left(\vec{a},t-\frac{R}{c} \right)
\left(\frac{|\vec{r}|-|\vec{r}-\vec{a}|}{c}\right)
\;dV\left(\vec{a}\right)\\
= &
\int_{\mathbb{B}\left(\vec{0},R_s\right)} \frac{\partial\vec{J}}{\partial t}\left(\vec{a},t-\frac{R}{c} \right)
\left(\frac{\vec{a}\cdot\hat{r}}{c} \right)
\;dV\left(\vec{a}\right) + O\left(\frac{R_s}{R}\right)\\
\end{align}
$$
where $\displaystyle \hat{r} = \frac{\vec{r}}{|\vec{r}|}$ is the unit vector in $\vec{r}$ direction.
Let $\displaystyle n_{\mu} = \frac{r_{\mu}}{R}$ be the components of the unit vector $\hat{r}$. Let $T_{\mu\nu}(t)$ be the $3 \times 3$ tensor defined by:
$$T_{\mu\nu}(t) = 
\frac{1}{c}\int_{\mathbb{B}\left(\vec{0},R_s\right)} \frac{\partial\vec{J}_{\mu}(\vec{a},t)}{\partial t} a_{\nu} \;dV\left(\vec{a}\right)
$$
Using Einstein summation convention, we have:
$$\left( \vec{U}(\vec{r},t) \times \hat{r} \right)_\mu 
=  \epsilon_{\mu\nu\omega} U_{\nu}(\vec{r},t) n_{\omega}
\sim  \epsilon_{\mu\nu\omega} T_{\nu\rho}(t-\frac{R}{c}) n_\rho n_{\omega}$$
where $\epsilon_{\mu\nu\omega}$ is the Levi-Civita symbol. Notice $d S(\vec{r}) = R^2 d\hat{r}$, the surface integral becomes:
$$\begin{align}
& \int_{\partial\mathbb{B}\left(\vec{0},R\right)} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \;dS\left(\vec{r}\right)\\
\sim & R^2 T_{\nu\rho} T_{\nu'\rho'} \left\{
\epsilon_{\mu\nu\omega} \epsilon_{\mu\nu'\omega'} 
\int n_{\rho} n_{\omega} n_{\rho'} n_{\omega'} d\hat{r}\right\}\\
= & R^2 T_{\nu\rho} T_{\nu'\rho'} \left\{
(\delta_{\nu\nu'}\delta_{\omega\omega'} - \delta_{\nu\omega'}\delta_{\nu'\omega})
\int n_{\rho} n_{\omega} n_{\rho'} n_{\omega'} d\hat{r}\right\}\\
= & R^2 \left\{ 
   T_{\nu\rho} T_{\nu\rho'} \int n_{\rho} n_{\rho'}  d\hat{r}
-  T_{\nu\rho} T_{\nu'\rho'} \int n_{\rho} n_{\nu'} n_{\rho'} n_{\nu} d\hat{r}
\right\}\\
= & R^2 \left\{ 
   \frac{4\pi}{3} T_{\nu\rho} T_{\nu\rho} 
-  \frac{4\pi}{15} ( T_{\nu\nu} T_{\rho\rho} + T_{\nu\rho} T_{\nu\rho} + T_{\nu\rho}T_{\rho\nu} )
\right\}\\
= & \frac{4\pi R^2}{15}\left\{4 T_{\nu\rho} T_{\nu\rho} - T_{\nu\nu} T_{\rho\rho} - T_{\nu\rho} T_{\rho\nu}
\right\}\tag{*}
\end{align}$$
where $\delta_{\mu\nu}$ is the Kronecker delta and we have used the identities:
$$\int n_{\mu} n_{\nu} d\hat{r} = \frac{4\pi}{3}\delta_{\mu}{\nu}
\quad\text{ and }\quad
\int n_{\mu} n_{\nu} n_{\omega} n_{\rho} d\hat{r} = \frac{4\pi}{15}
( \delta_{\mu\nu} \delta_{\omega\rho} + \delta_{\mu\omega}\delta_{\nu\rho} + \delta_{\mu\rho}\delta_{\nu\omega} )
$$
For general $\vec{J}$, your surface integral is most likely to grow like $R^2$.
Update
For the benefit of those not familiar with tensor, let me work out an important
special case using vector. If $\nabla\cdot\vec{J} = 0$, then one can use the fact
$\vec{J} = \vec{0}$ on $\partial\mathbb{B}(0,R_s)$ to show $T_{\mu\nu}$ is an antisymmetric
tensor. In particular, this implies one can find two vectors $\vec{A}$, $\vec{B}$ such that
$$T_{\mu\nu} = A_{\mu}B_{\nu} - B_{\mu}A_{\nu}$$
If one substitute this into $(*)$, the surface integral becomes:
$$\begin{align}
 & \frac{4\pi R^2}{15}\left(
4 ( A_{\nu}B_{\rho} - B_{\nu}A_{\rho})( A_{\nu}B_{\rho} - B_{\nu}A_{\rho} )
- ( A_{\nu}B_{\rho} - B_{\nu}A_{\rho})( A_{\rho}B_{\nu} - B_{\rho}A_{\nu} ) \right)\\
= & \frac{8\pi R^2}{3} \left( |\vec{A}|^2 |\vec{B}|^2 - (\vec{A}\cdot\vec{B})^2 \right)
= \frac{8\pi R^2}{3} |\vec{A} \times \vec{B} |^2
\end{align}$$
In terms of vector 
$$U_{\mu} = T_{\mu\nu} n_{\nu}
\quad\implies\quad
\vec{U}(\vec{r}) = \vec{A} (\vec{B}\cdot\hat{r} ) - \vec{B} (\vec{A}\cdot\hat{r})
= \hat{r} \times ( \vec{A} \times \vec{B} )
$$
and we can evaluate the surface integral as:
$$
\begin{align}
  & R^2 \int |( \hat{r} \times ( \vec{A} \times \vec{B} )) \times \hat{r}|^2 d\hat{r}
= R^2 \int \left( |\vec{A} \times \vec{B}|^2 - ( ( \vec{A} \times \vec{B} ) \cdot \hat{r} )^2 \right) d\hat{r}\\
= & 2\pi R^2 |\vec{A} \times \vec{B}|^2 \int_{0}^{\pi} (1 - \cos^2\theta) \sin\theta d\theta
= \frac{8\pi R^2}{3} |\vec{A} \times \vec{B}|^2
\end{align}$$
using suitable chosen polar coordinates. Of course, we get the same answer we obtained before using tensors.
For the more general case when $T_{\mu\nu}$ is not antisymmetric, computing the surface integral using vectors is actually more complicated and horrible than using tenors. I'll skip the mess here.
A: In addition to the full answer posted by @achillehui above, I'd like to add something that I should have thought of earlier (and in which case I wouldn't even have asked the question, as it turns out!)
We note that $\left|\vec{U}\left(\vec{r}\right) \times \hat{r}\right|^2$ is a scalar function or $\vec{r}$. So now if we choose a spherical coordinate system $\langle 0\le \mathrm{r} \lt \infty, \;0 \le \theta \le 2\pi,\; 0 \le \phi \le \pi\rangle$ with the origin at $\vec{0}$, we find that $$ds\left(\vec{r}\right) = \left(R\;\sin{\phi}\;d\phi\right)\left(R\;d\theta\right)$$ ... and since $R$ is a constant $\forall \vec{r} \in \partial\mathbb{B}\left(\vec{0},R\right)$, the the surface integral becomes:
$$
\oint_{\partial\mathbb{B}\left(\vec{0},R\right)} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2 \;ds\left(\vec{r}\right) 
= R^2 \int_0^{2\pi} \int_0^{\pi} \left|\vec{U}\left(\vec{r}\right) \times \hat{r} \right|^2  \sin \phi \;d\phi\;d\theta 
$$
So while the surface integral isn't proportional to $R^2$ (as the double integral is not a constant), it always contains $R^2$ as a factor. (This property was, in fact, all I needed for what I was doing.)
