# Flux integral of surface using Stokes theorem

1. May 19, 2013

### flaxstrax

1. The problem statement, all variables and given/known data
I have to control stokes theorem( I have to calculate line-and surface integral.
I have a vectorfield a=(3y,xz,yz^2).
And surface S is a paraboloid 2z=x^2+y^2. And it is limited by plane z=2.
For line integral the line is a circle C: x^2+y^2=4 on the plane z=2.
Vector n is pointing down ( Outwards).
[Broken]

2. Relevant equations
The line integral i got . it is -4*pi.
http://en.wikipedia.org/wiki/Stokes_theorem

3. The attempt at a solution
Okay so i get these wierd answers for surface integral and it should be -4 * pi.
I have done it in 5 different ways.
I find Flux a =(z^2-x,0,z-3) and then i find the n*dS. n*dS=-(f_x,f_y,1) = (x,y,-1) dx dy

Lets say that i use cylindrical coordinates , then x=r*cos t, y=r*sin t and z=z.
I find z = (x^2+y^2)/2 so it will be z=r^2/2
I think about the bonds , they should be 0 ≤ r ≤ 2 and 0 ≤ t ≤ 2*pi

Then i find the double integral →∬(z^2-x,0,z-3)*(x,y,-1) dx dy=
∬ z^2x-x^2-z+3 dS=
∬(r=0 to 2, t=0 to 2pi) r^4/4*r*r*cos t - r^2*r*(cos t)^2-r^2/2*r-3r dt dr =
∫(r=0 to 2) r^6/4 * sin t - r^3(t/2-1/2*cos t*sin t)-r^3/2*t+3rt dr=
∫(r=0 to 2) 0-r^3*pi - 0 - r^3* pi +6 r pi dr=
= ∫(r=0 to 2) -2*r^3* pi + 6*r*pi = -8*pi +12 * pi = 4 * pi

So i get 4 * pi from surface integral and -4*pi from line integral.
Can anyone explain what am i doing wrong ?

Last edited by a moderator: May 6, 2017
2. May 19, 2013

### CAF123

One thing that stands out is that in the diagram provided, the traversal of the curve is of the wrong orientation for applying Stokes'. This will generate an overall minus sign and could be the source of your error.

3. May 19, 2013

### flaxstrax

I dont understand what you mean ? Sorry, english is not my first language.

4. May 19, 2013

### CAF123

Actually, sorry the traversal is correct - ignore my last post.

5. May 19, 2013

### LCKurtz

It is correct to there. You have dropped a minus sign somewhere and trying to read that non-tex gives me eyestrain. The only term that should survive the $\theta$ integration is part of the $\cos^2\theta$ expression, and somewhere you lost the - in front of that term. It does come out $=-4\pi$.

Last edited by a moderator: May 6, 2017
6. May 19, 2013

### CAF123

The boundary of the surface S is the circle $x^2 + y^2 = 4$. However, this is also the boundary of the surface $x^2 + y^2 \leq 4$. Applying Stokes' to this surface rather than the paraboloid results in a much simpler integral.

7. May 19, 2013

### LCKurtz

Sure, but apparently the problem was to verify both sides of Stokes' theorem by calculating them directly.

8. May 19, 2013

### CAF123

Ah, okay, must have missed that.

9. May 19, 2013

### flaxstrax

So please check it out, i think u can understand it better this way :)

I dont see where i did wrong, and i cant really understand what you meant by " The only term that should survive the θ integration is part of the cos2θ expression, and somewhere you lost the - in front of that term." I have - in front ?

10. May 19, 2013

### LCKurtz

Sorry, but replacing one unreadable thing with another even more unreadable thing that takes forever to load isn't an improvement.

11. May 19, 2013

### flaxstrax

argh i havent learned how to write equations like you do ... And it takes a second for me to open that link , its HD pic of 1 A4 , like 2,3 mb or something. Ill try and write it then somehow.

12. May 19, 2013

### LCKurtz

I don't really need to see it. Did you find where you dropped the minus sign and do you now get the correct answer?

13. May 19, 2013

### flaxstrax

Is this better? No i don't see where...

$\int\int (z^2-x,0,z-3)*(x,y,-1) dxdy=$
$\int\int (r^4/4*r*cos t-r^2*(cos t)^2-r^2/2+3)r dt dr=$
$\int\int r^6/4*cos t+r^3*(cos t)^2-r^3/2+3r dt dr=$
$\int r^6/4*sin t-r^3(1/2(t+cos t*sin t)-r^3/2*t+3r*t dr=$
$\int 0-r^3*\pi-0-r^3*\pi+6r\pi dr=$
$\int -2*\pi*r^3+6*r*\pi dr=$
$-8\pi+12\pi=4\pi$

14. May 19, 2013

### LCKurtz

$$\int_0^2\int_0^{2\pi}\frac {r^6}{4}\cos(t) \color{red}{-}r^3\cos^2(t)-\frac{r^3}{2}+3r\,dtdr$$Like I suspected, you dropped a minus sign.

15. May 20, 2013

### flaxstrax

Yeah well that was just a typo, if i had calculated with + there, i would have the answer 12 * pi ,but it was just a typo.
Okay ill write the line integral too. But it says -4 pi is correct.

$r(t)=(2cos t,2 sin t, 2)$
$dr=(-2sin t,2cos t, 0 )$
$a=(3y,xz,yz^2)$
$I=\int (3y,xz,yz^2)(-2sin t,2cos t, 0 )dt=$
$\int (3y*-2*sin t+x*z*2*cos t)dt=$
$\int-12(sint)^2+8*(cos t)^2dt=$
$\int-12*(sint)^2+8-8*(sin t)^2dt=$
$16*\pi+\int-20(sin t)^2dt=$
$16*\pi-20*(1/2*t-cost*sint)=$
$16*\pi-20\pi=-4\pi$

16. May 20, 2013

### flaxstrax

Yeah well that was just a typo, if i had calculated with + there, i would have the answer 12 * pi ,but it was just a typo.
Okay ill write the line integral too. But it says -4 pi is correct.

$r(t)=(2cos t,2 sin t, 2)$
$dr=(-2sin t,2cos t, 0 )$
$a=(3y,xz,yz^2)$
$I=\int (3y,xz,yz^2)(-2sin t,2cos t, 0 )dt=($
$\int (3y*-2*sin t+x*z*2*cos t)dt=$
$\int-12(sint)^2+8*(cos t)^2dt=$
$\int-12*(sint)^2+8-8*(sin t)^2dt=$
$16*\pi+\int-20(sin t)^2dt=$
$16*\pi-20*(1/2*t-cost*sint)=$
$16*\pi-20\pi=-4\pi$

Sorry for double post , don't know why it happened.

17. May 20, 2013

### LCKurtz

You haven't put the limits in or showed enough that I can follow it, but are you doing$$\int_{2\pi}^0 -12\sin^2(t) + 8\cos^2(t)\, dt$$

That is the correct direction and that would change the sign if that isn't what you did. That would give $4\pi$ and explain why there doesn't seem to be anything wrong with the other calculation.