Evaluate $$\iint _{S} xz dx + yzdz + x^2 dz$$, where S is semisphere oriented from inside to outside with radius a centred at the origin and $z>0$.
So, basically i will do that: close the surface and apply the divergence. Let A be the semisphere and B be the disk of its intersection with z=0:
$$(\iint_{A}... = x) + \iint_{B} (xy,yz,x^2) * (0,0,-1) ds = \iiint 2z \space dV$$
$$x - \int_{0}^{2\pi} \int_{0}^{a} (r\cos(\phi))² r dr d\phi = \iiint 2r\cos(\theta) r^2 \sin(\theta)drd\theta d\phi$$
$$x - \frac{a^4 \pi}{4} = {\pi a^4}$$
So my answer is basically $$x = \frac{5 a^4 \pi}{4}$$
But the book answer is $$\frac{3 \pi a^4}{4}$$. The source of my possible error is evident, it occured in the third line. Apparenttly, it should be "$x + \frac{a^4 \pi}{4} = {\pi a^4}$". But i don't know why. Isn't my orientation right? I mean, all the normals are outside. Now i am really confused because i always thought that:
In the curl theorem, the orientations are such that the curve/surface are at left to the circulation. In the divergence theorem, the orientation are such that the surface has normal from inside to outside.
Isn't it right?
Note that
\begin{align} \int_0^{\frac\pi2} \int_0^{2\pi} \int_0^a 2r \cos (\phi)r^2 \sin (\phi) \,dr\, d\theta d \phi &=2\pi\cdot \frac{a^4}{4}\int_0^{\frac\pi2}\sin (2\phi) \, d\phi = \frac{\pi a^4}{2} \end{align}
Hence $$\frac{\pi a^4}{2}+\frac{\pi a^4}{4}=\frac{3\pi a^4}{4}$$