A question regarding flux of a vector field $\overrightarrow{F}$

So I have a surface: $$\text{S} : z=x^2+(y-R)^2$$

Vector field is defined by:

$$\overrightarrow{F}(x, y, z) = (y, - x, z(x^2+y^2))$$

I have to calculate flux in $$~2~$$ different scenarios.

First one is:

Through a part of the surface $$S$$ that has an additional condition: $$x^2+y^2 \leq R^2$$

So with this one I think, I should find the appropriate parametrization and just use the formula for surface integral:

$$\iint_{S} \overrightarrow{F} dP=\iint_{D} \overrightarrow{F}(\overrightarrow{r} (u, v) *(\overrightarrow{r_u} × \overrightarrow{r_v}) dudv$$

So I need help getting the correct parametrization to use, and the necessary bounds.

My idea is to manipulate the cylindrical coordinates into something that would work for my case. $$x=r\cos\phi$$ $$y=r\sin\phi$$ $$z=z$$ Would using $$z=x^2+(y-R)^2$$ and inserting x, y get me the correct parametrization? and just use $$r\leq|R|$$ and $$\phi\in[0,2\pi]$$ good enough, or am I doing something wrong?

Second one is through $$\delta G$$ where $$~G~$$ is bounded by $$S$$ from above and $$x^2+y^2=R^2 , z=0$$

So with this one I think of using Divergence theorem(Gauss theorem if you know it by this name) :

$$\iint_{\delta G} \overrightarrow{F} dP=\iiint_{G} div \overrightarrow{F} dV$$

so with this one I thought of using the cylindrical coordinates:

$$x=r\cos\phi$$ $$y=r\sin\phi$$ $$z=z$$ $$J=r$$

So i used the surfaces given to get:

$$r\in [0,R] , z\in [0, r^2-2\sin\phi + R^2]$$

I presume that $$[0,2\pi]$$ since there is no special conditions on it. Is this part okay, the way I got the bounds.

So any help would be appreciated.

I hope I did not put too many details in one question.

Thank you in advance.

For the first integral your approach is spot-on. But I suggest first using the parametrization \begin{align} & x = x \\ & y = y \\ & z = x^2 + (y-R)^2 \\ \end{align} because without calculation you can always use that $$r_x \times r_y = (-f_x,-f_y,1)$$ After you take the dot product, $$\mathbf{\text{then}}$$ convert the scalar double integral into polar coordinates. This method saves a lot of time and heartache from the use of potential obscure trig identities.

Your second part looks alright, but your notation is extremely unfamiliar to me, so I have a few questions.

When you put (a) and (b) next to your surfaces, are they denoting parts (a) and (b) of a question or just two surfaces to bound the region with?

And when you go into cylindrical coordinates, what is J?

Lastly, your $$z$$ upper bound should be $$r^2 - 2Rr\sin \phi +R^2$$

• i just used the a) amd b) to show where i got bounds – MathIsTheWayOfLife Aug 29 '19 at 20:27
• what about the answer to my other question? – Ninad Munshi Aug 29 '19 at 20:38
• oh you mean $J$, that is the Jacobi determinant, we use this notation, since it's divergence theorem, it is the regular triple integral, so Jacobi comes in use – MathIsTheWayOfLife Aug 29 '19 at 20:41
• Oh ok, the way you had it formatted I was concerned it was another variable. Then your second approach works too. Try my suggestion for the first question and you'll have the best method for both problems. – Ninad Munshi Aug 29 '19 at 20:43