calculusvector analysis
When applying Gauss' Theorem to a sphere centered at the origin with radius 1 for example, the boundaries would be $\phi: 0 \;$ to $\; \pi$, $\theta : 0 \;$ to $\; 2\pi$, and $\rho: 0 \;$ to $\; 1$.
What if the sphere was centered at $(2,3,5)$ for example with the same radius, would the boundaries here change?
The divergence theorem says that
$\int_{\partial M} \vec X \cdot \vec n dS = \int_M \nabla \cdot \vec X dV, \tag{1}$
where $M$ is a closed, bounded region in $\Bbb R^3$ with smooth boundary $\partial M$ which is an orientable surface in $\Bbb R^3$ with outward pointing normal field $\vec n$; here $dV$ is the volume element in $M$ and $dS$ is the area element on $\partial M$. The flux of $\vec X$ through $\partial M$ is, by definition, the left-hand integral in (1). In the present case, it is particulary easy to evaluate the volume integral on the right-hand side of (1); taking $\vec X = F = 3z^2 i + 2y j + xk$, we see that
$\nabla \cdot F = \dfrac{\partial(3z^2)}{\partial x} + \dfrac{\partial(2y)}{\partial y} + \dfrac{\partial(x)}{\partial z} = 2; \tag{2}$
the flux of $F$ through the unit sphere $S^2$ is then, by (1),
$\int_{S^2} F \cdot \vec n dS = \int_B 2 dV = 2 \int_B dV, \tag{3}$
where $B = \{(x, y, z) \in \Bbb R^3 \mid x^2 + y^2 +z^2 \le 1 \}$ is the unit ball. We have
$\int_B dV = \dfrac{4 \pi}{3}, \tag{4}$
so that
$\int_{S^2} F \cdot \vec n dS = \dfrac{8 \pi}{3}. \tag{5}$
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
There are several conflicting conventions for spherical polar coordinates among mathematicians, physicists, geographers and astronomists. In this answer, I will stick to following parametrization of the sphere:
$$[0,2\pi] \times \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \ni (\theta,\phi) \quad\mapsto\quad (x,y,z) = (R\cos\phi\cos\theta,\,R\cos\phi\sin\theta,\,R\sin\phi ) \in \mathbb{R}^3$$
i.e. $\theta, \phi$ corresponds to the longitudes and latitudes in geography and $\phi \in [0,\frac{\pi}{2}]$ corresponds to the northern hemisphere.
The question at hand is find a curve whose tangent is making an constant angle $\alpha$ with the circles of latitudes. In above parametrization, the metric on the sphere is given by $$ds^2 = R^2 (\cos^2\phi \, d\theta^2 + d\phi^2)$$
What this means is:
if we move along the line $\phi = \text{const}$ for a small amount $\delta\theta$, the distance traveled $\approx R \cos\phi \delta\theta$.
if we move along the line $\theta = const$ for a small amount $\delta\phi$, the distance traveled $\approx R \delta\phi$.
In order for the curve to make a angle $\alpha$ with the circles of latitudes, we need to make displacement $(\delta\theta, \delta\phi)$ such that
$$\frac{R \delta\phi}{R \cos\phi \, d\theta}\approx m \stackrel{\text{def}}{=} \tan\alpha$$
This implies the curve satisfies following ODE:
$$\frac{1}{\cos\phi}\frac{d\phi}{d\theta} = m \quad\iff\quad\frac{d\sin\phi}{1 - \sin^2\phi} = m \, d\theta $$
If the curve start at a point $(\theta_0,\phi_0)$, we can solve above ODE and get: $$\frac{1 - \sin\phi}{1 + \sin\phi} = \frac{1 - \sin\phi_0}{1 + \sin\phi_0} e^{-2m(\theta - \theta_0)} $$ For example, if we start the curve at $(\theta_0,\phi_0) = (0,0) \iff (x,y,z) = (R,0,0)$, we get $$ \begin{align} \frac{1-\sin\phi}{1+\sin\phi} = e^{-2m\theta} &\iff \sin\phi = \tanh(m \theta)\\ &\iff \begin{cases} x &= R \frac{\cos\theta}{\cosh(m\theta)}\\ y &= R \frac{\sin\theta}{\cosh(m\theta)}\\ z &= R \tanh(m\theta) \end{cases} \end{align} $$ Such curves are called Rhumb line, there are more details on its wiki entry.
At the end is a picture of $6$ rhumb lines. All of them start at $(\theta_0,\phi_0) = (0,0)$ and their $\alpha$ vary from $10^\circ$ (the red curve) to $60^\circ$ (the cyan curve). As one can see, they sort of "spiral" around the north pole as they approach it.
$\hspace0.8in$ 
Best Answer
They would not change, but the parametrization of the sphere would change. It would be:
$$(x,y,z) = (2 + \rho \cos \theta \cos \phi, 3 + \rho \sin \theta \cos \phi, 5 + \rho \sin \phi)$$
Note also that the Jacobian of the parametrization wouldn't change, which was something expected, as the sphere doesn't change shape and is only translated to another place.