Let S be the lower half of the sphere x^2 + y^2 + z^2 = 2, oriented toward the inner side. Compute the surface integral of (x^2 + xz) dy dz + (y^2 + yz) dz dx + (2z^2 + 1) dx dy.

Step-by-step solution

1. 1

   Read it as a flux. The integral is $\iint_\Sigma\mathbf{F}\cdot d\mathbf{S}$ for $\mathbf{F}=(x^2+xz,\;y^2+yz,\;2z^2+1)$. Since $\Sigma$ is not closed, the divergence theorem needs a cap.

2. 2

   Close the surface with the flat disk. Add $D:\{x^2+y^2\le 2,\ z=0\}$. Together $\Sigma$ and $D$ bound the lower half-ball $V$. On the closed boundary the outward normal points down/outward on $\Sigma$ and upward ($+\mathbf{k}$) on $D$.

3. 3

   Apply Gauss. $\nabla\cdot\mathbf{F}=(2x+z)+(2y+z)+4z=2x+2y+6z$. By symmetry of $V$ in $x$ and in $y$, the $2x$ and $2y$ pieces integrate to zero, leaving $\iint_{\partial V,\text{out}}\mathbf{F}\cdot d\mathbf{S}=6\iiint_V z\,dV.$

4. 4

   Evaluate the remaining triple integral. In spherical coordinates with $0\le\rho\le\sqrt2$, $\tfrac\pi2\le\varphi\le\pi$, $\iiint_V z\,dV=2\pi\int_{\pi/2}^{\pi}\cos\varphi\sin\varphi\,d\varphi\int_0^{\sqrt2}\rho^3\,d\rho=2\pi\left(-\frac12\right)(1)=-\pi,$ so the closed outward flux is $6(-\pi)=-6\pi$.

5. 5

   Subtract the disk. On $D$ we have $z=0$, so $dy\,dz$ and $dz\,dx$ contribute nothing and the third component is $2(0)^2+1=1$. The upward flux is just the disk area: $\pi(\sqrt2)^2=2\pi$. Hence the outward flux through $\Sigma$ is $-6\pi-2\pi=-8\pi$.

6. 6

   Flip to the inner side. The problem asks for the inner orientation, which reverses the normal: $\iint_{\Sigma,\text{inner}}\mathbf{F}\cdot d\mathbf{S}=-(-8\pi)=8\pi.$ So the blank is filled with $8$.