<s>[INST] Using the divergence theorem to calculate the surface integral of a sphere Hi so I have the question to use the divergence theorem to calculate the surface integral of the sphere

Let $S=\{(x,y,z):(x-a)^2 + (y-b)^2 + (z-c)^2 = R^2\}$ and $f = (x^2,y^2,z^2)$. Evaluate
  $$I=\iint_S fdS.$$ 

By the divergence theorem, I get that
$$I=\iiint_B \text{div}(f)dV$$
where $B=\{(x,y,z):(x-a)^2 + (y-b)^2 + (z-c)^2 \leq R^2\}$.
I get $\text{div}(f) = 2x+2y+2z$, I’m just wondering where to go from here, are the $(x-a)^2$ and $0$ the boundaries for my triple integral?
 [/INST] We have to evaluate
$$\iiint_{(x-a)^2 + (y-b)^2 + (z-c)^2 \leq R^2} (2x+2y+2z)dxdydz$$
which is equal to
$$2\iiint_{X^2 + Y^2 + Z^2 \leq R^2} ((a+X)+(b+Y)+(c+Z))dXdYdZ$$
where $X=x-a$, $Y=y-b$, $Z=z-c$. 
 Since the ball $X^2 + Y^2 + Z^2 \leq R^2$ is symmetric with respect to the plane $X=0$, it follows that the integral of $X$ over the ball is zero. The same for $Y$ and $Z$.  Therefore, we finally obtain 
$$2(a+b+c)\cdot \frac{4\pi R^2}{3}.$$
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