Find the flux for the vector field $F =(x^2,y^2,z^2)$ Find the flux for the vector field $\vec{F}=(x^2,y^2,z^2)$ across the boundary to the ball given as $$(x-1)^2 +(y+1)^2 +(z-2)^2 \le 4$$
Edit: How much I've done: The center of the sphere is $(1,−1,2)$ (I've sketched it), thus $r(θ,ϕ)=(1+cosθsinϕ, −1+sinθsinϕ, 2+cosϕ)$ . Then to find the normal vector $\vec{N}$ , I've calculated $\frac{∂r(θ,ϕ)}{∂ϕ}=(cosθcosϕ,sinθcosϕ,−sinϕ)$ and $\frac {∂r(θ,ϕ)}{∂θ}=(−sinθsinϕ,cosθsinϕ,0)$ then $\frac{∂r(θ,ϕ)}{∂ϕ} \times \frac{∂r(θ,ϕ)}{∂θ}$, and finally got $(sin2ϕ,sinθsin2ϕ,sinϕcosϕ)$. Am I on the right track?
 A: The Divergence Theorem is absolutely the right approach, as all the hypotheses of the theorem hold. You should always prefer to do a volume integral rather than a surface integral. In this case, you might also recognize that
\begin{multline*}
\iiint_K x\,dV = \bar x\, \text{volume}(K), \quad \iiint_K y\,dV = \bar y\, \text{volume}(K), \quad\text{and} \\ \iiint_K z\,dV = \bar z\, \text{volume}(K),
\end{multline*}
where $(\bar x,\bar y,\bar z)$ is the centroid (center of mass) of $K$.
When $K$ is a ball (with uniform density), we all know that its centroid is at its center. Indeed, I would comment that multivariable calculus students should always be on the lookout to apply symmetry arguments.
A: Your region is $R: (x-1)^2+(y+1)^2+(z-2)^2 \leq  4$.
If you are applying divergence theorem, you need to evaluate $\displaystyle \iiint_R 2(x+y+z) \ dV$
Just to simplify your working, consider the below substitution -
$u = x-1, v = y+1, w = z-2$ and region $R$ transforms to $E: u^2+v^2+w^2 \leq 4$ which is a sphere of radius $2$ centered at the origin. Jacobian for the transformation is $1$.
So the integral becomes $\displaystyle \iiint_E 2(2 + u + v + w) \ dV$. Note that $u, v, w$ are odd functions and due to symmetry, their integral over a sphere centered at the origin will be zero. For example, integral of $u$ over half sphere (for $u \leq 0$) will cancel out with the integral over the other half ($u \geq 0)$.
So the problem reduces to finding $\displaystyle \iiint_E 4 \ dV$ which you know even without the integral.
