Calculate $\int_V\frac{1}{2+(x^2+y^2+z^2)^{\frac{3}{2}}}dxdydz$, where $V=\{(x,y,z)|x^2+y^2+z^2\lt 1\}$ using the co area formula.
So I know that the formula is: $\int_V f dx$ = $\int_a^b\int_{M_c}\frac{f(x)}{|\nabla \phi(x)|}dS dc$, where $\phi:V\rightarrow\mathbb{R}\in C^1,\forall x: \nabla \phi(x)\neq 0,\phi(V)=(a,b)$ and $M_c=\{x=(x_1,...,x_n)\in V|\phi(x)=c,c\in(a,b)\}$.
My attempt -
So the difficult part here (for me) is to find the correct $\phi$. I thought about $\phi = x^2+y^2+z^2$ which is $\phi \in(0,1)$ but $\nabla \phi(0,0,0) =0$ so I cannot use it.
I also thought about $\phi = (x^2+y^2+z^2)^{\frac{3}{2}}$, but also here $\nabla \phi(0,0,0) =0$ so I cannot use it.
Other attempt was to use the function that we integrate - $\phi = \frac{1}{2+(x^2+y^2+z^2)^{\frac{3}{2}}}$ and here I know that $\nabla \phi \neq 0 \forall x$, but when I lower the dimension of the integral, which variable should be gone?(i.e., instead of $3$ dimensional, $dxdydz$, what should it be?)
Or perhaps, is there any better $\phi$ to use in this case?
Hint. If you pick
$$ \phi(\mathbf{x}) = \sqrt{x^2+y^2+z^2} $$
one has $\parallel \nabla \phi \parallel =1$ and the coarea formula becomes the formula for change of variables into spherical coordinates.