# Flux on a sphere above the plane $z=0$ (Surface is not closed)

Given the vector field $$F=((y-1)^2+z^2,(z^2-x^2),(y-1)^2)$$ , Let $$S$$ be the surface of the sphere $$x^2+y^2+z^2-2y=0$$ which is above $$xy$$ meaning above the plane $$z=0$$ ' let $$n$$ be the unit vector normal outward the sphere ( pointing up) , show that the flux in the given direction is $$0.25\pi$$ $$\iint_SF \cdot n dS=0.25\pi$$ ( we need to solve using gauss and divergence theorem)

My try: First I organized what is given:

our sphere and plane looks like this

our field is $$F=((y-1)^2+z^2,(z^2-x^2),(y-1)^2)$$ and we have two equations $$\begin{cases} x^2+(y-1)^2+z^2=1\\ z=0 \end{cases}$$ since we need to solve using gauss and divergence theorem we have - $$\iint_SF \cdot n dS=\iiint_V div(F) \cdot n dxdydz$$

It is obvious that $$div(F)=0$$ , and we can also see that $$S$$ does not enclose the volume $$V$$ so we need to close it and on this we can apply the theorem.

we can close it with $$D$$ the projection on $$Z=0$$ and we get $$\iint_S F \cdot ndS +\iint_D F \cdot ndD = \iiint_V div(F) \cdot n dV$$ since the $$div(F)=0$$ we get that $$\iint_S F \cdot ndS =- \iint_D F \cdot ndD$$

Here I got stuck , I tried to calculate $$\iint_D F \cdot ndD$$ but I keep getting the wrong answer , assuming $$D$$ is the projection and according to the functions we have I got that it is the circle $$x^2+(y-1)^2=1$$ I tried using polar coordiantes where $$0 \leq \theta \leq \pi$$ and $$0 \leq r \leq 1$$ and I kept getting a different answer than $$0.25 \pi$$.

Can anyone give me any tips and hints? I want to understand this topic more it is interesting but I am having a hard time with it.

Thank you and sorry for the English mistakes hope The translations are understandable.

The spherical surface is $$S1: x^2 + (y-1)^2 + z^2 = 1, z \geq 0$$

Now applying divergence theorem, you correctly concluded that the net flux through the closed surface is zero. The disk that we use at $$z= 0$$ to close the surface is $$S2: x^2 + (y-1)^2 \leq 1$$.

We can parametrize $$S2$$ as $$\ x = r \cos\theta, y = 1 + r\sin\theta, 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$$.

The outward normal vector to the disk is $$(0, 0, -1)$$ as it is in plane $$z = 0$$.

So $$\ \vec F \cdot \hat n = - (y-1)^2 = - r^2 \sin^2\theta$$

The integral to find flux through the disk is,

$$\displaystyle \int_0^{2\pi} \int_0^1 - r^3 \sin^2\theta \ dr \ d\theta = - \frac{\pi}{4}$$

We subtract this from the net flux through the closed surface to get flux through the spherical surface $$S1$$. Hence the answer is $$\displaystyle \frac{\pi}{4}$$.

• Because we are parametrizing the disk as $x = r \cos\theta, y = 1 + r\sin\theta$. So essentially we measure $\theta$ considering the center $(0, 1)$ as origin. If we use $x = r\cos\theta, y = r\sin\theta$ then yes we would have $0 \leq \theta \leq \pi$. May 27 at 17:52
• In that case $x^2 + y^2 - 2y = 0$ will be written as $r^2 - 2 r \sin\theta = 0 \implies r = 2 \sin\theta$. So $0 \leq r \leq 2\sin\theta, 0 \leq \theta \leq \pi$. May 27 at 17:54
• Need an inequality describing $S_2$. May 27 at 17:59
• @Aruralreader thank you May 27 at 18:00
• @Adamrk one last point - The parametrization I used in this answer works for any circle whereas the other one I mentioned in comments works only for circle with center on one of the coordinate axes and the other coordinate axis being tangent to the circle. Both parametrizations are important to learn and use. Sometimes one is easier vs. the other. In this case given $(y-1)^2$ in vector field, the one used in the answer is easier. May 27 at 18:01