Verify Divergence Theorem

Question

Verify the Divergence Theorem when the vector-field F has the form,

$\mathbf{F}(x,y,z) =x^2\hat{\imath}+z\hat{\jmath}$

and D is the upper half-ball $x^2+y^2+z^2≤ 1, z ≥ 0$

I am having trouble doing so, as I achieve two different answers.
When solving it by $\iint F\cdot ndS$ I get an answer, $\frac{-π}{4}$

and when solving it by $\iiint \nabla\cdot F dxdydz$ I get an answer of, $\frac{2π}{3}$

I would want to attach my work but I am not familiar with LaTeX and it would take me ages.

I would greatly appreciate it anyone to guide me through the answer.

Answer 1

Note that you cannot apply Gaus-Ostrogradski theorem (Divergence theorem) on a non - compact surface.

Meaning we need surface K=$\{(x,y,0)| x^2+y^2\le 1\}$

Lets try the first value. But first we need the normal $\vec{n}$

$$ z_x'=-\frac{x}{\sqrt{1-x^2-y^2}}\\ z_y'=-\frac{y}{\sqrt{1-x^2-y^2}} $$

Meaning $$ \vec{n}=\left (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1 \right ) $$ Where $z=\sqrt{1-x^2-y^2}$

$$ \iint_S x^2 dx +zdy +0 dz=\iint_D \frac{x^3}{\sqrt{1-x^2-y^2}}+y\,\,dxdy=0 $$

And in the case when we apply Gaus-Ostrogradski on the Upper ball surface and K.

We get

$$ \iiint_T 2x \,\,\, dxdydz=2\int_{0}^{1}\int_{-\pi}^{\pi}\int_{0}^{\pi/2} \rho^2 \sin(\theta)\cos(\sigma) d\theta d\sigma d\rho=\frac{2}{3}\int_{-\pi}^{\pi}\cos(\sigma)d\sigma $$

Now we just need to prove that $\iint_K \vec{F} \cdot (dx,dy,dz)=0$

$$ \iint_K x^2 dx + 0 dy + 0dz=\iint_D \left (x^2,0,0 \right )\cdot \left( 0,0,1 \right) dxdy=0 $$

We have now proven the equality.

Answer 2

$\nabla F=2x$

In this case, it's easier to use spherical coordinates. So we have $0<\rho<1, 0<\theta<2\pi, 0<\phi< \frac{\pi}{2}$. $\iiint \nabla F dxdydz = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 2\rho*sin(\phi)*cos(\theta)\;d\rho\;d\theta\;d\phi=0$

And if you calculate the flux,

$z=(1-y^2-x^2)^{\frac{1}{2}}$

$z_{x}=-(\frac{x}{1-y^2-x^2})^{\frac{1}{2}}$

$z_{y}=-(\frac{y}{1-y^2-x^2})^{\frac{1}{2}}$

$\iint F*ndS= \iint F(x,y,x)*(-z_{x},-z_{y},1) dS=0$