Expert: Paul Klarreich - 12/24/2006

Question
Hi - I'm confused with how to solve this problem :
Compute the flux integral [double integral] G(x,y,z) dot ndA for the vector field G(xyz) = 1i+2j+4x^2k through the circular region of radius 3 in the yz plane, centered on the x axis and oriented in the positve x direction. Some hints provided: integral sin(theta) cos(theta) d(theta) = (sin^2theta)/2  AND cos^2phi = (1+cos(2phi))/2

Thank You for your help!

Answer
Questioner:  Melody
Category:  Advanced Math
Subject:  Multivariable Calculus
Question:  Hi - I'm confused with how to solve this problem :
Compute the flux integral [double integral] G(x,y,z) dot ndA for the vector field G(xyz) = 1i+2j+4x^2k through the circular region of radius 3 in the yz plane, centered on the x axis and oriented in the positve x direction. Some hints provided: integral sin(theta) cos(theta) d(theta) = (sin^2theta)/2  AND cos^2phi = (1+cos(2phi))/2

Thank You for your help!
.........................................
Hi, Melody,

I'm sorry, but it has been a long time since I did this and I don't think I ever taught it, either.  But that has never stopped me from doing silly things before, so.....

As I recall, the flux integral refers to some 'stuff' moving through a surface in a particular direction, called, I suppose, 'out'.  (As opposed to 'in'.)

So you want the component of the flow that is normal to the surface, and you integrate it over the entire surface area.  

Your flux integral is:

{{
|| G(x,y,z) * N dA
}}

where G = i + 2j + 4x^2 k

and the integral is over the circle of radius 3 in the yz plane, oriented in the positive x-direction.
............................
Now I can't say I really understand this, BUT:

If you have a circle (or any flat surface) in the y-z plane, then wouldn't the normal to that surface be in the x-direction?  That is, your unit normal, N, should be just:

1i + 0j + 0k.

(Positive 1, because you did say 'positive x-direction', right?)

And your component in that direction is:

G * N = (i + 2j + 4x^2 k)(1i + 0j + 0k) = 1.

Then your integral is nothing more than:

{{
|| dA
}}

over that circle.  And that is just its area, 9pi.

Of course, I never used your 'hints', which are standard integration methods, which is a bit worrisome.

But that's it for what it's worth.  This answer is provided as is, with no warranty, express or implied.  If it's wrong, no refund will be made.