# Thread: another flux through surface question

1. ## another flux through surface question

Compute the flux of the vector field, $\displaystyle \vec{F}$, through the surface, S. $\displaystyle \vec{F} = y\vec{i} + 7\vec{j} - xz\vec{k}$ and S is the surface $\displaystyle y = x^2 + z^2$ with $\displaystyle x^2 + z^2 \leq 36$ oriented in the positive y direction.

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

I don't know if this is correct...

2. 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

3. 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;

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

and I don't know how to represent this in polar

4. yes the ingrl is okay

in polar form just think of z as y

x= rcos(t)

z = r sin(t)

5. hmm.. so then it becomes:

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

am I missing something here?

6. I was using t for theta --just lazy

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

7. 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??

8. yes since x = rcos(theta) you'll end up integrating r^4 first

where does -432 come from?

9. 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?

10. No r is a variable you take into account that r is 6 when you integrate r from 0 to 6

11. shouldn't we multiply it by r first and then take the integration with respect to r?

12. 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

13. well after that I got:

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

is this correct??

14. Yes

15. Originally Posted by Calculus26
Yes
moving on forward I have:

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

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

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