continuity of function defined by surface integral I'm considering this function for positive $r$
$$
\varphi(r) = \int\limits_{\partial B(0,r)} f \, dS.
$$
where $f$ is a $C^1(\mathbb{R}^n)$ function and $\partial B(0,r)$ is the surface of the $n$-sphere.
I think this function should be continuous (even if $f$ is only continuous) but I am having a hard time proving it, I should find a small enough $h>0$ so 
$$
\left| \partial B(0,r+h) - \partial B(0,r) \right| < \varepsilon.
$$
My failed ideas
-This would be routine if I had $\partial B(0,r) \subset \partial B(0,r+h)$ but that is false, however if I could somehow transform the integral from the boundary to the interior of the ball that would solve it but I don't know if that's possible.
-The other idea was to show that the $\varphi(r) = f(c_r) \operatorname{Measure}(\partial B(0,r))$ for a $c_r$ depending on r and going from there, but then I need assume that $\lim\limits_{h \to 0} c_{r+h} = c_r$ but that's pretty much assuming what I'm trying to prove.

I'm probably missing something important here, so any hints would be greatly appreciated.
 A: I think you can use the higher dimensional version of divergence theorem.  
First, turns $f$ into a vector valued function $\vec F$ which has direction always pointing radically outwards.
Then $f\;dS=\vec F\cdot \vec n \; dS$ where $\vec n$ is the normal vector.  
So now by divergence theorem $\displaystyle\int\limits_{{B(0,r)}}\nabla\cdot\vec F\;dV=\int\limits_{\partial B(0,r)}\vec F\cdot \vec n\;dS$
Since $f$ is $C_1$, we can do the divergence.
Now $B(0,r)\subset B(0,r+h)$.  
Does this help?
A: I think you can do something like the following: let $\varepsilon,\varepsilon'>0$ be given. Then we may choose $h$ so small that $$\phi(r+h)=\int_{\partial B(0,r+h)}f(x)\mathrm{d}S=\int_{\partial B(0,r)}f(x(r+h)/r)\mathrm{d}S+O(\varepsilon') \int_{\partial B(0,r+h)}|f(x)|\mathrm{d}S,$$ the error term appearing because we changed the measure. The last integral being finite, we may just absorb this into the $O(\varepsilon'),$ and by choosing $\varepsilon'$ small enough (and for correspondingly small $h$), we may ensure that $O(\varepsilon')<\varepsilon.$ We also have $$\left|\int_{\partial B(0,r)}f(x)-f(x(r+h)/r)\mathrm{d}S\right|\leq\int_{\partial B(0,r)}|f(x)-f(x(r+h)/r)|\mathrm{d}S,$$ and once $h$ is small enough, $|f(x)-f(x(r+h)/r)|\leq\max\{\|\triangledown f(x)\|: x\in\partial B(0,r)\}(r/h)<\varepsilon,$ since this maximum must be finite on a compact set (and using the fact that $f\in C^{1}(\mathbb{R}^{n})$). That means the integral above is bounded by $\varepsilon\cdot\mathrm{meas}(\partial B(0,r)).$ Putting it all together, we have (for sufficiently small $h$): \begin{align*}|\phi(r)-\phi(r+h)|&=\left|\int_{\partial B(0,r)}f(x)-f(x(r+h)/r)\mathrm{d}S\right|+O(\varepsilon')\\&< \varepsilon(\mathrm{meas}(\partial B(0,r))+1).\end{align*} Since our choice of $\varepsilon$ was arbitrary, $\phi$ is continuous.
A: Assume that we are working in spherical coordinates. Then you can rewrite your integral as 
$$g(r) =\int_{\partial B_r}f=\int_{\frac{- \pi}{2}}^{\frac{\pi}{2}}\int_0^{2 \pi} f(\theta, \phi, r) d\theta d\phi$$
Since $f$ is globally $C^1$ you can differentiate under the integral sign (with respect to $r$), so $g$ is differentiable, therefore continuous.
