Let S be the surface formed by the part of the paraboloid $z = 1- x^2-y^2$ lying above the $xy$-plane and let $\vec F= x\hat i + y\hat j+2(1-z) \hat k$.
Calculate the flux of $\vec F$ across S, taking the upward direction as the one for which the flux is positive.
Do this in two ways:
a) By direct calculation of flux by $\iint_s \vec F .\hat n \;dS$.
b) By computing the flux of $\vec F$ across a simpler surface and using the divergence theorem.
I am quite new to multi-variable ,so please bear with me.
The problem i am having is for part a) , however i have tried part b) as follow.
I know that, $\text{div}\; \vec F=0$ .
Hence we can imagine a imaginary surface $x^2+y^2\leq1$ at $z=0$ with normal vector $ \hat k$ And the surface S already present. Combining them to create a closed surface through which flux will be zero as.
$\iiint \text{div}\; \vec F . dV=F_1+F_2$ where, $F_1$ is considered as flux through as Paraboloid surface $S$ and $F_2$ is through the circular disc described.
So $F_1=-F_2$.
Now, flux through circular disc is inward hence negative.
$F_2=-\iint \vec F . \hat k \; dA$
gives
$F_2=-\iint 2 . dA= -2\pi$ and $F_1=2 \pi$.
For part a), Here is what i tried, I tried to evaluate this in polar form, firstly i find normal in cartesian to paraboloid, which is given as.
$\hat n = \frac{2x \hat i + 2y \hat j + \hat k}{\sqrt{1+4x^2+4y^2}}$ .
Computing $\vec F . \hat n = \frac{2x^2+2y^2+2(1-z)}{\sqrt{1+4x^2+4y^2}}=\frac{4(1-z)}{\sqrt{5-4z}}$ , since all this flux is evaluated on $S$.
Now converting this to polar coordinates in 3D $<r\cos \theta \sin \phi , r \sin \theta \sin \phi , r \cos \phi>$.
and $dS=r^2 \sin \phi d\theta d\phi$.
Gives $F_1=\iint_{s}\frac{4(1-r \cos \phi)}{\sqrt{5-4r\cos \phi}}. r^2 \sin \phi d \theta d \phi $.
I tried to solve for $r$ in terms of $\phi$ giving $r=\frac{2}{\sqrt{1+3\sin^2 \phi}+\cos \phi}$ (after removing discontinuity and solving for $r$ by equation of paraboloid, taking positive root.)
Add: Here's how i did this, take the equation of paraboloid we get, $r^2 \sin^2 \phi = 1- r\cos \phi$, And solving quadratic equation in terms of $r$. Taking the positive root.
Since $r$ is independent of $\theta$ our integral becomes.
$F_1= 8 \pi \int_{0}^{\pi/2}\frac{1-r \cos \phi}{\sqrt{5-4r \cos \phi}}\times r^2 \sin \phi d\phi $.
This is where i get stuck, I have no idea from here on.
If this is relevant, http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/exams/prac4b.pdf
Best Answer
The problem is when you do $$\vec{F}\cdot \vec{n}$$ Your integral should become $$\iint \vec{F}\cdot\vec{n}dS =\iint\vec{F}\cdot \frac{\nabla \vec{f}}{||\nabla \vec{f}||} ||\nabla \vec{f}|| dA =\iint 4(1-z)dA$$
You should not have the denominator when you use polar coordinate.