Use Gauss's Law to calculate the triple integral , why two ways two answers? $\iiint_V (xy+yz+zx) \mathrm{d}x\mathrm{d}y\mathrm{d}z$
$V: x\geq 0 , y \geq 0 , 0\leq z \leq 1, x^2+y^2\leq 1$
The triple integral is $\frac{11}{24}$
The first way:
\begin{array}{l}
=\frac{1}{2} \iint_{S}\left(x^{2} y \mathrm{d} y \mathrm{d} z+y^{2} z \mathrm{d} z \mathrm{d} x+z^{2} x \mathrm{d} x \mathrm{d} y\right) \\
=\frac{1}{2}\left[\iint_{D_{y z}}\left(1-y^{2}\right) y \mathrm{d} y \mathrm{d} z+\iint_{D_{z x}}\left(1-x^{2}\right) z \mathrm{d} z \mathrm{d} x+\iint_{D_{x y}} x \mathrm{d} x \mathrm{d} y\right] \\
=\frac{1}{2}\left[\int_{0}^{1} \mathrm{d} y \int_{0}^{1}\left(1-y^{2}\right) y \mathrm{d} z+\int_{0}^{1} \mathrm{d} x \int_{0}^{1}\left(1-x^{2}\right) z \mathrm{d} z+\int_{0}^{1} x \mathrm{d} x \int_{0}^{\sqrt{1-x^{2}}} \mathrm{d} y\right] \\
=\frac{1}{2}\left[\int_{0}^{1}\left(1-y^{2}\right) y \mathrm{d} y+\frac{1}{2} \int_{0}^{1}\left(1-x^{2}\right) \mathrm{d} x+\int_{0}^{1} x \sqrt{1-x^{2}} \mathrm{d} x\right] \\
=\frac{11}{24}
\end{array}
The second way:
\begin{array}{l}
=\frac{1}{2} \iint_{S}\left(x y z\mathrm{d} y \mathrm{d} z+x y z \mathrm{d} z \mathrm{d} x+xyz \mathrm{d} x \mathrm{d} y\right) \\
=\iint_{D_{y z}}\left(1-y^{2}\right) y z \mathrm{d} y \mathrm{d} z+\iint_{D_{z x}} x\left(1-x^{2}\right)  z \mathrm{d} z \mathrm{d} x+\iint_{D_{x y}} xy \mathrm{d} x \mathrm{d} y \\
= 2\int_0^1\mathrm{d} z \int_0^1 (1-y^2)yz \mathrm{d} y+\int_0^1\mathrm{d}x\int_{0}^{\sqrt{1-x^2}} xy \mathrm{d} y\\
=2\times\frac{1}{2}\times(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}\times(\frac{1}{2}-\frac{1}{4})\\
=\frac{3}{8}
\end{array}
 A: Note that
\begin{equation}
\iiint_V \left(xy+yz+zx \right)dxdydz=\iiint_V \nabla\cdot\left(xyz,xyz,xyz\right) dxdydz=\iint_{\partial V} \left(xyz,xyz,xyz\right)\cdot n_{\partial V} d\text{vol}_{\partial V},
\end{equation}
where we used Gauss' Theorem, where $n_{\partial V}$ is the outward unit normal on the boundary of your domain.
Now the rest is a simple calculation.
A: $ \displaystyle \iiint_V (xy+yz+zx) ~ dx ~ dy ~ dz$
where $V: x^2 + y^2 \leq 1, x, y \geq 0, 0 \leq z \leq 1$
The answer is indeed $\frac{11}{24}$. Now you want to evaluate it by surface integral using a vector field $\vec F$ such that,
$ \nabla \cdot \vec F = xy + yz + zx$
Divergence theorem gives you flux for an outward vector field over a closed region. So the surface we need to evaluate integral over is $S: S_1 \cup S_2 \cup S_3 \cup S_4 \cup S_5$. $S_1$ is the cylindrical surface in the first quadrant. $S_2$ is quarter disc at $z = 0$, $S_3$ is quarter disc at $z = 1$, $S_4$ is rectangular surface at $x = 0$ and $S_5$ is rectangular surface at $y = 0$.
Surface integral is $ \displaystyle \iint_S \vec F \cdot \hat n ~ dS$.
$ \vec F = (xyz, zyz, xyz)$
Normal vector to $S_1$ is $(x, y, 0)$. Normal vector to $S_3$ is $(0, 0, 1)$. Integral over $S_2, S_4$ and $S_5$ is zero.
Parametrizing $S_1$ in cylindrical coordinates as,
$x = \cos\theta, y = \sin\theta, z = z$
$ \displaystyle \iint_{S_1} \vec F \cdot \hat n ~ dS  = \displaystyle \int_0^{\pi/2} \int_0^1 z(\cos^2\theta \sin\theta + \sin^2\theta \
cos \theta) ~ dz ~ d\theta $
Parametrizing $S_3$  as
$x = r \cos\theta, y = r \sin\theta, z = 1$
$ \displaystyle \iint_{S_3} \vec F \cdot \hat n ~ dS  = \displaystyle \int_0^{\pi/2} \int_0^1 r^3 \cos\theta \sin\theta ~ dr ~ d\theta $
Adding both $S_1$ and $S_3$ leads to the answer of $\frac{11}{24}$.
You can apply the same approach for the first way too.
