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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

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  • $\begingroup$ You'll probably be better off with cylindrical coordinates: sphericals are normally a pain unless you actually have spheres involved. $\endgroup$ – Chappers Jun 4 '15 at 17:03
  • $\begingroup$ Is my part b) correct though? I'd try cylindrical thank you. $\endgroup$ – Mann Jun 4 '15 at 17:05
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    $\begingroup$ b) looks okay, although it's better to talk about the direction the normal vector points (the question says the normal on the paraboloid points upwards, so the one on the disc should point down to be out of the enclosed volume). This is one of the things you have to be careful about explaining, or no one knows what you're doing: I had to do some guesstimate calculations to make sure I believed you... $\endgroup$ – Chappers Jun 4 '15 at 17:13
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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.

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  • $\begingroup$ So my integral is, $8 \pi \int_{0}^{\pi/2} (1-r\cos \phi) r^2 \sin \phi d\phi $ ? And thank you. $\endgroup$ – Mann Jun 4 '15 at 17:18
  • $\begingroup$ @Mann: At this point, it would be easier to replace $z$ by $1-x^2-y^2$ and use 2D polar coordinates instead of 3D spherical coordinates. $\endgroup$ – KittyL Jun 4 '15 at 17:19
  • $\begingroup$ Sorry for my bad question, I have no skill with this, I am starting to learn but , how would i define my $dA$ in 2D then? Isn't the surface 3D.*(Hope i didn't say something stupid.)* $\endgroup$ – Mann Jun 4 '15 at 17:23
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    $\begingroup$ @Mann: Considering the surface as parametrized as $z=1-x^2-y^2$, in terms of $x,y$. So the $dA$ is the region of the surface projected onto $xy$ plane, which is the disc $x^2+y^2\leq 1$. This is a double integral then. So you can use polar coordinates. $\endgroup$ – KittyL Jun 4 '15 at 17:56
  • $\begingroup$ Ah thank you, i got it now! I surely had bit confusions with this stuff but this made it clear! $\endgroup$ – Mann Jun 4 '15 at 18:02

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