flux integral.....argh!!!

**swinburneguy2009** · May 12th 2010, 09:24 PM

Hi all, I am having real trouble with this problem. If anyone can steer me in the right direction, or better yet, help by showing the solution, I would be much appreciated:

Consider a surface

S which is that portion of the plane 2x + 2y + z = 6

included in the first octant.
Evaluate the flux integral ∫∫ F nˆdA , where the vector field

F = [xy,−x^2 , x + z] and nˆ is a unit outer normal vector to S.

Thanks in advance!

**HallsofIvy** · May 13th 2010, 03:30 AM

I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.

Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (6- 2x- 2y)\vec{j}$ .

The derivatives with respect to x and y are $\vec{r}_x= \vec{i}- 2\vec{j}$ and $\vec{r}_y= \vec{j}- 2\vec{k}$ . Those vectors are in the plane and their cross product, $2\vec{i}+ 2\vec{j}+ \vec{k}$ is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is the "vector differential of area" for the plane Ax+ By+ Cz= D.

Now, what about the orientation? You say "where $\vec{n}$ is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?

If positive z direction, take the dot product of $\vec{F}= xy\vec{i}- x^2\vec{j}+ (x+ y)\vec{k}$ with $2\vec{i}+ 2\vec{j}+ \vec{k}$ and integrate with respect to x and y. If the negative z direction, take the dot product of $\vec{F}$ with $-2\vec{i}- 2\vec{j}- vec{k}$ and integrate (that will just change the sign on the answer).

The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.

**AllanCuz** · May 17th 2010, 08:57 PM

Originally Posted by HallsofIvy

I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.

Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" $\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (6- 2x- 2y)\vec{j}$ .

The derivatives with respect to x and y are $\vec{r}_x= \vec{i}- 2\vec{j}$ and $\vec{r}_y= \vec{j}- 2\vec{k}$ . Those vectors are in the plane and their cross product, $2\vec{i}+ 2\vec{j}+ \vec{k}$ is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector $A\vec{i}+ B\vec{j}+ C\vec{k}$ is the "vector differential of area" for the plane Ax+ By+ Cz= D.

Now, what about the orientation? You say "where $\vec{n}$ is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?

If positive z direction, take the dot product of $\vec{F}= xy\vec{i}- x^2\vec{j}+ (x+ y)\vec{k}$ with $2\vec{i}+ 2\vec{j}+ \vec{k}$ and integrate with respect to x and y. If the negative z direction, take the dot product of $\vec{F}$ with $-2\vec{i}- 2\vec{j}- vec{k}$ and integrate (that will just change the sign on the answer).

The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.

Would we not be better served by going to the definitions directly? Because we have a plane we will have 2 flux integrals to compute: the top and bottem. Of course you address this in your post (actually, everything i'm about to say you've already addressed) but I think for the purposes of the student it's just easier to learn how to "plug and chug" so to speak (i know...i know this doesn't actually facilitate learning the material, but if it helps understand how to at least attempt these problems, why not?)

Let us write all the things we know,

Our function is defined as,

$F = xy \hat i - x^2 \hat j + (x+z) \hat k$

Our surface is defined as,

$z = 6-2x-2y$

And for the computation of our integral,

$\hat N dS = +/- ( - \frac{ \partial S }{ \partial x } \hat i - \frac{ \partial S }{ \partial y } \hat j + k )$

Notice how we have a plus and a minus, this means we need to find both of these to get to the total flux! If you just want upward or downward, only compute the flux integral for either of those.

$\hat N dS = +/- ( + 2 \hat i + 2 \hat j + 1k )$

Thus,

$Total Flux = \iint_D F \cdot ( + 2 \hat i + 2 \hat j + 1k ) dA + \iint_D F \cdot ( - 2 \hat i - 2 \hat j - 1k ) dA$

To get our bounds we let $z = 0$ which yields $x + y = 3$

And,

$0 \le x \le 3$ with $0 \le y \le 3 - x$

At this point everything should be clear enough for you to go ahead with this problem.

HallsofIvy gave a much more mathetmatical explanation then myself, but this should allow you to tackle the problem in a brute force manner.

**11rdc11** · May 17th 2010, 09:20 PM

Originally Posted by swinburneguy2009

Hi all, I am having real trouble with this problem. If anyone can steer me in the right direction, or better yet, help by showing the solution, I would be much appreciated:

Consider a surface

S which is that portion of the plane 2x + 2y + z = 6

included in the first octant.
Evaluate the flux integral ∫∫ F nˆdA , where the vector field

F = [xy,−x^2 , x + z] and nˆ is a unit outer normal vector to S.

Thanks in advance!

I would tell you work this Stoke's Theorem. It would be easier to integrate the curl

**HallsofIvy** · May 18th 2010, 02:35 AM

Originally Posted by AllanCuz

Would we not be better served by going to the definitions directly? Because we have a plane we will have 2 flux integrals to compute: the top and bottem.

No, you don't.

Of course you address this in your post (actually, everything i'm about to say you've already addressed) but I think for the purposes of the student it's just easier to learn how to "plug and chug" so to speak (i know...i know this doesn't actually facilitate learning the material, but if it helps understand how to at least attempt these problems, why not?)

Let us write all the things we know,

Our function is defined as,

$F = xy \hat i - x^2 \hat j + (x+z) \hat k$

Our surface is defined as,

$z = 6-2x-2y$

And for the computation of our integral,

$\hat N dS = +/- ( - \frac{ \partial S }{ \partial x } \hat i - \frac{ \partial S }{ \partial y } \hat j + k )$

Notice how we have a plus and a minus, this means we need to find both of these to get to the total flux! If you just want upward or downward, only compute the flux integral for either of those.

No, calculating the flux through a surface in both directions will sum to 0. That's why Swinburneguy2009 said "outer normal to S" although, as I pointed out "outer" doesn't really apply to a plane. The "upward" flux will be the negative of the "downward" flux- if you calculate both and total them, you will get 0.

$\hat N dS = +/- ( + 2 \hat i + 2 \hat j + 1k )$

Thus,

$Total Flux = \iint_D F \cdot ( + 2 \hat i + 2 \hat j + 1k ) dA + \iint_D F \cdot ( - 2 \hat i - 2 \hat j - 1k ) dA$

To get our bounds we let $z = 0$ which yields $x + y = 3$

And,

$0 \le x \le 3$ with $0 \le y \le 3 - x$

At this point everything should be clear enough for you to go ahead with this problem.

HallsofIvy gave a much more mathetmatical explanation then myself, but this should allow you to tackle the problem in a brute force manner.

**AllanCuz** · May 18th 2010, 03:51 AM

Originally Posted by HallsofIvy

No, you don't.

No, calculating the flux through a surface in both directions will sum to 0. That's why Swinburneguy2009 said "outer normal to S" although, as I pointed out "outer" doesn't really apply to a plane. The "upward" flux will be the negative of the "downward" flux- if you calculate both and total them, you will get 0.

I understand this, the main point of my post was to get across we dont need to use positional vectors. I know the total flux out of a plane is equal to 0, but I was showing him how to set it up for future problems and noted that he can pick/choose which direction he wants.

It's easiar to work with the given definitions then to go back to the basics in this case.

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