# another flux through surface question

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• Apr 21st 2009, 06:09 PM
TheRekz
another flux through surface question
Compute the flux of the vector field, $\vec{F}$, through the surface, S. $\vec{F} = y\vec{i} + 7\vec{j} - xz\vec{k}$ and S is the surface $y = x^2 + z^2$ with $x^2 + z^2 \leq 36$ oriented in the positive y direction.

$\int\limits_R ((x^2+z^2)\vec{i} + 7\vec{j} - xz\vec{k}) \cdot (-2x\vec{i} - 2z\vec{k} + \vec{j}) dA$

I don't know if this is correct...
• Apr 21st 2009, 07:08 PM
Calculus26
it looks good-however the orientatation seems a little strange.

you have a parabaloid aligned along the positive y axis

Usually we specify inward or outward--here on the half in the first and 8th octants you have an outward normal but in the 2d and 5th you have an inward normal.

So is this correct?

and of course your region of integration is a circle of radius 6 in the x-z plane
• Apr 22nd 2009, 08:59 AM
TheRekz
Quote:

Originally Posted by TheRekz
so is it okay?? I don't know what the limits of integral is..

reason I ask is because I get;

$\int\limits_R (-2x^3+7) dA$

and I don't know how to represent this in polar
• Apr 22nd 2009, 11:15 AM
Calculus26
yes the ingrl is okay

in polar form just think of z as y

x= rcos(t)

z = r sin(t)
• Apr 25th 2009, 11:30 PM
TheRekz
hmm.. so then it becomes:

$\int_0^{2pi}\int_0^6 (-432(cos(t)^3)+7) rdrd\theta$

am I missing something here?
• Apr 26th 2009, 12:31 PM
Calculus26
I was using t for theta --just lazy

integral of cos(theta)^3 should not present a pblm
• Apr 26th 2009, 12:52 PM
TheRekz
Quote:

Originally Posted by Calculus26
I was using t for theta --just lazy

integral of cos(theta)^3 should not present a pblm

yes but i do need to multiply it by r??
• Apr 26th 2009, 12:56 PM
Calculus26
yes since x = rcos(theta) you'll end up integrating r^4 first

where does -432 come from?
• Apr 26th 2009, 01:04 PM
TheRekz
Quote:

Originally Posted by Calculus26
yes since x = rcos(theta) you'll end up integrating r^4 first

where does -432 come from?

because r is equal to 6, and therefore :

x = rcos(theta)
= (6cos(theta))^3
= 216cos(theta)^3

multiplying that by -2, that's where I got the -432cos(theta)^3, am I doing something wrong here? so it should be -432cos(theta)^3r, right?
• Apr 26th 2009, 01:07 PM
Calculus26
No r is a variable you take into account that r is 6 when you integrate r from 0 to 6
• Apr 26th 2009, 01:47 PM
TheRekz
shouldn't we multiply it by r first and then take the integration with respect to r?
• Apr 26th 2009, 01:51 PM
Calculus26
again x= rcos(theta) when you integrate over the disk of radius 6

r varies from 0 to 6 it isn't a constant 6

so you'd have (-2 r^4 cos^3(theta) +7) drd(theta) to integrate
• Apr 26th 2009, 01:58 PM
TheRekz
well after that I got:

$\int_0^{2\pi}\frac{-15552(cos^3\theta)}{5}+42 d\theta$

is this correct??
• Apr 26th 2009, 02:02 PM
Calculus26
Yes
• Apr 26th 2009, 02:06 PM
TheRekz
Quote:

Originally Posted by Calculus26
Yes

moving on forward I have:

$\frac{15552(sin(\theta)^3-3sin(\theta))}{15} +42\theta$

evaluated from 0 to 2pi.. resulting in 84pi and webassign doesn't take this answer
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