# Calculate integral where S is the surface of the half ball $x^2 + y^2 + z^2 = 1, \space z \geq 0,$ and $F = (x + 3y^5)i + (y + 10xz)j + (z - xy)k$

I am asked to calculate this integral and want to make sure I am doing this set up correctly, So I tried to paramatize this surface by :

$$x = rcos(\theta), \space y = rsin(\theta), \space z = \sqrt{1 - r}$$

$$0 \leq r \leq 1$$ and $$0 \leq \theta \leq 2\pi$$

then I found :

$$\Phi_{\theta} = -rsin(\theta)i + rcos(\theta)j + 0k , \space \Phi_{r} = cos(\theta)i + sin(\theta)j + \frac{1}{2\sqrt{1-r}}k$$

which would make:

$$\Phi_{\theta} \times \Phi_{r} = \frac{1}{2}\bigl(\frac{rcos(\theta)}{\sqrt{1-r}}\bigr)i + \frac{1}{2}\bigl(\frac{rsin(\theta)}{\sqrt{1-r}}\bigr)j -rk$$

then my surface integral would be :

$$\int_0^{2\pi} \int_0^1 ((rcos(\theta) + 3(rsin(\theta))^5)i + (rsin(\theta) + 10(rcos(\theta))(\sqrt{1 - r}))j + (\sqrt{1 - r} - (rcos(\theta)rsin(\theta))k )\space \cdot (\frac{1}{2}\bigl(\frac{rcos(\theta)}{\sqrt{1-r}}\bigr)i + \frac{1}{2}\bigl(\frac{rsin(\theta)}{\sqrt{1-r}}\bigr)j -rk )$$

is this set up correct ? I have tried to reduce it after taking the dot product with the trig identity's but I still end up with a fairly complicated expression

• It should be $z = \sqrt{1-r^2}$, not $\sqrt{1-r}$. – Nick Nov 21 '18 at 23:17

## 1 Answer

The problem is ambiguous. If the surface of the half ball includes the flat surface at $$z=0$$ the problem simplifies to a very simple integral using the divergence theorem. If not, the problem simplifies too but needs to calculate the flux through that flat surface.

1.- For the flat surface included ($$B$$ is the half ball, $$\partial B$$ its surface and $$\mathbf F=F_x\mathbf i+F_y\mathbf j+F_z\mathbf k$$), the divergence theorem gives:

$$\phi=\int_{\partial B}\mathbf F·\mathbb d\mathbf S=\int_B\nabla·\mathbf F\,\mathbb dV$$

$$\nabla·\mathbf F=\dfrac{\partial F_x}{\partial x}+\dfrac{\partial F_y}{\partial y}+\dfrac{\partial F_z}{\partial z}=1+1+1=3$$

In spherical coordinates $$(r,\theta,\phi)$$

$$\phi=\int_B 3\mathbb d V=\int_0^1\int_0^{\pi/2}\int_o^{2\pi}3r^2\sin\theta\,\mathbb d\phi\,\mathbb d\theta\,\mathbb d r=2\pi$$

2.- If the flat surface, $$D=\{x^2+y^2\leq 1;z=0\}$$, is not included we need to subtract the flux through it:

For the surface element with its vector pointing outwards is $$-\mathbf k$$:

$$\mathbb d\mathbf S=-\mathbb dx\mathbb dy\,\mathbf k$$

For the flux through the surface element ($$z=0$$):

$$\mathbf F·\mathbb d\mathbf S=((x + 3y^5)\mathbf i + (y + 10xz)\mathbf j + (z - xy)\mathbf k)·(-\mathbb dx\mathbb dy\,\mathbf k)=xy\,\mathbb dx\mathbb dy$$

$$\phi_D=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}xy\,\mathbb dx\mathbb dy=0$$

The flux is zero. It seems that in any case the answer is $$2\pi$$