Surface Integral - or Line Integral?

[bon]

Homework Statement

Air is flowing with a speed of 0.4m/s in the direction of the vector (-1, -1, 1). Calculate the volume of air flowing per second through the loop which consists of straight lines joining, in turn, the following (1,1,0), (1,0,0), (0,0,0), (0,1,1), (1,1,1) and (1,1,0).

Homework Equations

The Attempt at a Solution

So I don't know if this is meant to be a line integral or surface one?

My feeling is that it should be a surface integral over that pentagonal surface..I.e. double integral of F.n dS..where F is (-1,-1,1)..

Firstly, is this right? Secondly, how do I use the fact that the speed of flow is 0.4? Thirdly, how do I find the normal to the plane?! Finally, how do i integrate over the surface of the pentagon in the double integral?!

Thanks!

 

[tiny-tim]

Hi bon!

Hint: it's not a plane!

(draw a cube, and then trace the line around the cube)

 

[bon]

Ah okay thanks - so do i use divergence theorem to simplify the surface integral over the cube?

How do I use the speed of the flow?

THanks!

 

[tiny-tim]

It might be simplest in this case to just shade in those parts of the faces of the cube that the line crosses, and calculate the flow for each shaded face separately.

 

[joe:)]

Im trying to do a similar problem to this..

As bon says though, I can't see how you use the speed being 0.4m/s here?!

What I would do is dot (-1,-1,1) with n hat for and integrate over two surfaces separately - one with corners (1,1,0), (1,0,0) and (1,1,1)

and then over the other surface..

But how do you use the speed on the vector field?

 

[joe:)]

So summing those two integrals I get..2-1/2 = 1.5

But as you say bon, I'm not sure how to use the speed being 0.4m/s

 

[tiny-tim]

Im trying to do a similar problem to this..

As bon says though, I can't see how you use the speed being 0.4m/s here?!

Hi joe:)!

volume per second = length x area per second = speed x area

or, more precisely, = ∫ velocity "dot" normal d(area)

 

[joe:)]

Hi joe:)!

volume per second = length x area per second = speed x area

or, more precisely, = ∫ velocity "dot" normal d(area)

Very helpful. Thanks!

So I need to find a velocity vector in the direction (-1,-1,1) with magnitude 0.4?

So is it 0.4/root3 (-1,-1,1) then do I just dot this with the two normals for the two surfaces and carry out the two double integrals as I did..?

Was I right in getting 1.5? In which case i guess the actual answer should be 4/5root3 - 0.4/2root3 = root3/5? Correct?

THANK YOU :)

 

[tiny-tim]

Hi joe:)!

(have a square-root: √ )

Very helpful. Thanks!

So I need to find a velocity vector in the direction (-1,-1,1) with magnitude 0.4?

So is it 0.4/root3 (-1,-1,1) then do I just dot this with the two normals for the two surfaces and carry out the two double integrals as I did..?

That's right!

(but I haven't checked your actual figures)

 

[joe:)]

Thanks tiny-tim :)