Find the volume between bounded by $z=4$ and $z=x^2+y^2$.(Answer: $8\pi$)
I thouhg using dievergence theorm ($\iint_KdivFdxdydz=\iint_SF\cdot\hat{n}dS$) for $\vec{F}=\big(\frac x 2,\frac y 2,4\big)$. suppose we devide into $K_1=\{(x,y,z)\mid z=4\}$ and $K_2=\{(x,y,z)\mid 0\le x^2+y^2\le4\},$ On $K_1$ the integral is (suppose the normal is $(0,0,1)$) $4\cdot \pi\cdot 2=8\pi$. Allegedly the integral on $K_2$ had to be 0 but for the parameterization of $K_2$, $\varphi(r,\theta)=(r\cos\theta,r\sin\theta,r^2)$ where $r\in[0,2],\theta\in[0,\frac \pi 2]$ but $$\varphi_r^\prime\times\varphi_\theta^\prime=(-2r^2\cos\theta,-2r^2\sin^2\theta,r)$$so $$||\varphi_r\times\varphi_\theta||=\sqrt{4r^4(\cos^2\theta+\ \sin^2\theta)+r^2}=r\sqrt{4r^2+1}$$ and getting normalized normal ($\hat n=\big(-\frac{2r\cos\theta}{\sqrt{4r^2+1}},-\frac{2r\sin\theta}{\sqrt{4r^2+1}},\frac{1}{\sqrt{4r^2+1}}\big)$),$$\vec F\cdot \hat n=\frac{-2r(\cos^2\theta+\sin^2\theta)+4}{\sqrt{4r^2+1}}$$ and it follows that $$ \iint_{K_2}F\cdot\hat n=\int_0^2\int_0^\frac \pi 2\frac{4-2r}{\sqrt{4r^2+1}}d\theta dr\thickapprox 0.6\neq 0$$ Where am I wrong?
As before, we divide the boundary $S$ into the two pieces, $S_1=\{(x,y,z)| (z=4) \wedge (x^2+y^2 \le 4)\}$ and $S_2=\{(x,y,z)| (z\le4) \wedge (x^2+y^2 = z) \}.$
We would like $\vec{F} \cdot \hat{n} = 0$ on $S_2$. The symmetry of the paraboloid suggests we want each $\vec{F}$ to lie in a common plane with the $z$ axis, and to be tangent to the parabola that is the intersection of $K_2$ and that plane. In fact that parabola has equation $z = r^2$ where $r = \sqrt{x^2 + y^2}$. This suggests $F_z = 2rF_r$ where $F_r$ is the radial component of $\vec{F}$ in cylindrical coordinates.
So let's try $$\vec{F}=\left(\frac x 4,\frac y 4, \frac z 2\right).$$ Then $\nabla \cdot \vec{F} = 1$, so that the integral of $\nabla \cdot \vec{F}$ over the solid region $K$ bounded by $S$ gives the volume of that region. And anywhere on $S_2$,
$$ \begin{eqnarray} {\sqrt{4r^2+1}} (\vec{F} \cdot \hat{n}) &=& (-2r\cos\theta)\frac x 4 + (-2r\sin\theta)\frac y 4 + \frac z 2 \\ &=& -\frac{x^2}{2} - \frac{y^2}{2} + \frac{x^2+y^2}{2} \\ &=& 0. \end{eqnarray} $$
Now, $S_1$ is a circular disk of radius $2$, so its area is $4\pi$. Everywhere on $S_1$, we have $z = 4$, and so $$\vec{F} \cdot \hat{n} = F_z = \frac z 2 = 2.$$ Integrating this over $S_1$ gives us $2\cdot(4\pi) = 8\pi$.