1. Divergence Theorem

Using the divergence theorem, evaluate

$\int {\int}_\Sigma F.nds$,

where

$F = (4x, -2y^2, z^2), \Sigma$ is the bundary of the region defined by $x^2 + y^2 \le 4, 0\le z \le 3$, and n is the outward unit normal.

examples in notes only do this then F differentiates so that once the first integration is done, it can be simplified to something times $r = x^2 + y^2$, if you follow me. This one doesnt do that as far i can see so i dont know what to do...

My workings so far...

$\nabla F = 4 - 4y + 2z^2$

$\Rightarrow \int {\int}_\Sigma F.nds = \int \int {\int}_R (4 - 4y + 2z^2)dxdydz$

$= \int {\int}_D (12 - 12y + 18)dxdy$

$= \int {\int}_D (30 - 12y)dxdy$

Now is the point i was talking about where it has become simplified so that r may be used but i have no idea what to do here.

If it helps heres the latex i used so you dont have to type it out again...
\int \int {\int}_R (4 - 4y + 2z^2)dxdydz = $\int \int {\int}_R (4 - 4y + 2z^2)dxdydz$

just modify that

Cheers!

Anyone else here from Edinburgh and pissed off about spending the last week in the library revising for the exams while outside had turned into the mediterranean and now come the weekend... Suns gone away... Grim.

Using the divergence theorem, evaluate

$\int {\int}_\Sigma F.nds$,

where

$F = (4x, -2y^2, z^2), \Sigma$ is the bundary of the region defined by $x^2 + y^2 \le 4, 0\le z \le 3$, and n is the outward unit normal.

examples in notes only do this then F differentiates so that once the first integration is done, it can be simplified to something times $r = x^2 + y^2$, if you follow me. This one doesnt do that as far i can see so i dont know what to do...

My workings so far...

$\nabla F = 4 - 4y + 2z^2$

$\Rightarrow \int {\int}_\Sigma F.nds = \int \int {\int}_R (4 - 4y + 2z^2)dxdydz$

$= \int {\int}_D (12 - 12y + 18)dxdy$

$= \int {\int}_D (30 - 12y)dxdy$

Now is the point i was talking about where it has become simplified so that r may be used but i have no idea what to do here.

[snip]
I'll assume you're correct to here:

$\int \int_D (30 - 12y) \, dx \, dy$.

Now evaluate this double integral by transforming to polar coordinates:

$x = r \cos \theta$, $y = r \sin \theta$ and $dx \, dy = r \, dr \, d\theta$ (r is the Jacobian of the transformation).

Then your double integral becomes $\int_{\theta = 0}^{\theta = 2 \pi} \int_{r = 0}^{r = 2} (30 - 12 r \sin \theta) r \, dr \, d \theta$.

Note: Given that your closed surface forms a cylinder, you could have cut to the chase and switched to cylindrical polar coordinates to do the volume integral.

3. Hi,

I too get stuck at the polar coordinates point and so have a few questions:
1) what if the shape of the boundary is not a circle (i.e. is a sphere, cylinder etc...), then what happens, do the polar coordinates change?
2) how to you determine the direction of n?
3) how do you determine the boundaries of r?

thanks!

4. Originally Posted by pyrotechnics
Hi,

I too get stuck at the polar coordinates point and so have a few questions:
1) what if the shape of the boundary is not a circle (i.e. is a sphere, cylinder etc...), then what happens, do the polar coordinates change?
A sphere and cylinder are not two dimensional so "polar coordinates", which are, would not apply. If the boundary were a cylinder, then you would use "cylindrical coordinates" which are just polar coordinates with an added "z" variable. If the boundary were a sphere, you would use spherical coordinates.

2) how to you determine the direction of n?
n is always perpendicular to the boundary- whether it is "in" or "out", "left" or "right", etc. depends upon the [b]orientation[b] of the boundary. If you are integrating counter-clockwise around a closed boundary, you would take the normal vector that points inward. If you are integrating clockwise around the boundary, you would take the normal vector that points outward. Here you were told "n is the outward unit normal" which means that you are integrating around the curve clockwise. In general, imagine that you are [b]walking[b] around the boundary facing in the direction in which you are integrating. Your left arm extends in the direction of the normal vector.

3) how do you determine the boundaries of r?
I assume you mean the limits of integration for r. Imagine a line extending from (0, 0) to the boundary. If (0, 0) is inside the boundary, the limits will be from 0 to the value of r where that line intersects the boundary. If (0, 0) is outside the boundary, that ine will intersect the boundary twice. The limits of integration will be from the value of r at the first intersection to the value of r at the second.

thanks!

Using the divergence theorem, evaluate

$\int {\int}_\Sigma F.nds$,

where

$F = (4x, -2y^2, z^2), \Sigma$ is the bundary of the region defined by $x^2 + y^2 \le 4, 0\le z \le 3$, and n is the outward unit normal.

examples in notes only do this then F differentiates so that once the first integration is done, it can be simplified to something times $r = x^2 + y^2$, if you follow me. This one doesnt do that as far i can see so i dont know what to do...

My workings so far...

$\nabla F = 4 - 4y + 2z^2$

$\Rightarrow \int {\int}_\Sigma F.nds = \int \int {\int}_R (4 - 4y + 2z^2)dxdydz$

$= \int {\int}_D (12 - 12y + 18)dxdy$

$= \int {\int}_D (30 - 12y)dxdy$

Now is the point i was talking about where it has become simplified so that r may be used but i have no idea what to do here.

If it helps heres the latex i used so you dont have to type it out again...
\int \int {\int}_R (4 - 4y + 2z^2)dxdydz = $\int \int {\int}_R (4 - 4y + 2z^2)dxdydz$

just modify that

Cheers!

Anyone else here from Edinburgh and pissed off about spending the last week in the library revising for the exams while outside had turned into the mediterranean and now come the weekend... Suns gone away... Grim.

$\nabla F = 4 - 4y + 2z^2$

it should be

$\nabla F =4 - 4y +2z$

6. Originally Posted by 11rdc11
$\nabla F = 4 - 4y + 2z^2$
$\nabla F =4 - 4y +2z$