Flux integral of surface using Stokes theorem

[flaxstrax]

Homework Statement

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).

Homework Equations

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

The Attempt at a Solution

Okay so i get these weird 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 ?

 

[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.

 

[flaxstrax]

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.

I don't understand what you mean ? Sorry, english is not my first language.

 

[CAF123]

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

 

[LCKurtz]

Homework Statement

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).

Homework Equations

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

The Attempt at a Solution

Okay so i get these weird 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=

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 $=-4\pi$.

 

[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.

 

[LCKurtz]

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.

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

 

[CAF123]

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

Ah, okay, must have missed that.

 

[flaxstrax]

http://www.upload.ee/image/3322664/20130520_011715.jpg

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

I don't see where i did wrong, and i can't 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 ?

 

[LCKurtz]

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

 

[flaxstrax]

argh i haven't 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.

 

[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?

 

[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

 

[LCKurtz]

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=

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

 

[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

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

 

[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

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.

 

[LCKurtz]

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=

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.