Calculate this integral using Stokes

[supercali]

Homework Statement

let F be vector field:
\[\vec F = \cos (xyz)\hat j + (\cos (xyz) - 2x)\hat k\]
let L be the the curve that intersects between the cylinder (x - 1)^2 + (y - 2)^2 = 4<br /> and the plane y+z=3/2
calculate:
\[\left| {\int {\vec Fd\vec r} } \right|\]

Homework Equations

in order to solve this i thought of using the stokes theorem because the normal to the plane is \[\frac{1}{{\sqrt 2 }}(0,1,1)\]
thus giving me
\oint{Fdr}=\int\int{curl(F)*n*ds}=\int\int{2\sqrt{2}*\sin(xyz)}


i tried to parametries x y and z x= rcos(t)+1 y=rsin(t)+2 z=1/2-rsin(t)

but it won't work this that sin...

 

[NoMoreExams]

I don't know the rules but you should probably stick to posting the same question in one place only?

 

[HallsofIvy]

Homework Statement

let F be vector field:
\[\vec F = \cos (xyz)\hat j + (\cos (xyz) - 2x)\hat k\]
let L be the the curve that intersects between the cylinder (x - 1)^2 + (y - 2)^2 = 4<br /> and the plane y+z=3/2
calculate:
\[\left| {\int {\vec Fd\vec r} } \right|\]

Homework Equations

in order to solve this i thought of using the stokes theorem because the normal to the plane is \[\frac{1}{{\sqrt 2 }}(0,1,1)\]
thus giving me
\oint{Fdr}=\int\int{curl(F)*n*ds}=\int\int{2\sqrt{2}*\sin(xyz)}


i tried to parametries x y and z x= rcos(t)+1 y=rsin(t)+2 z=1/2-rsin(t)

but it won't work this that sin...

The cylinder has radius 2 so you should have x= 2cos(t)+ 1, y= 2sin(t)+ 2, z= 3/2- y= -1/2- 2sin(t).

However, it is not clear to me what you are doing! You said you you wanted to use Stokes theorem so you should be integrating over the surface, not the boundary. The surface that curve bounds is given by y+ z= 3/2 so, using x and y as parameters, the "vector differential of area" d\vec{S}= \vec{n}dS= (\vec{j}+ \vec{k})dxdy

I get \nabla\times F d\vec{S}= 2 dxdy.

 

[NoMoreExams]

Same discussion here: https://www.physicsforums.com/showthread.php?t=251286 which is why I think it's a good idea to only start the thread in one place. Just my .02

 

[supercali]

The cylinder has radius 2 so you should have x= 2cos(t)+ 1, y= 2sin(t)+ 2, z= 3/2- y= -1/2- 2sin(t).

However, it is not clear to me what you are doing! You said you you wanted to use Stokes theorem so you should be integrating over the surface, not the boundary. The surface that curve bounds is given by y+ z= 3/2 so, using x and y as parameters, the "vector differential of area" d\vec{S}= \vec{n}dS= (\vec{j}+ \vec{k})dxdy

I get \nabla\times F d\vec{S}= 2 dxdy.

first of all sorry for posting twice it was by a mistake somthing got stuck and i didnt notice that it was already posted...

secondly Hall i kind of lost you here :"using x and y as parameters, the "vector differential of area" i didnt really understand that i do get it that the plane z+y=3/2 is the surface and that it is bounded by the cylinder but when claculating the curl(F) i just don't get it how does it disapear the sin(xyz)?

 

[HallsofIvy]

If \vec F = \cos (xyz)\vec j + (\cos (xyz) - 2x)\vec k
Then \nabla\times \vec{F}= (xysin(xyz)-xzsin(xyz))\vec{i}+ (yzsin(xyz)+ 2)\vec{j}- yzsin(xyz)\vec{k}

The dot product of that with \vec{j}+ \vec{k}, a normal vector to y+ z= 3/2, is 0(xysin(xyz)-xzsin(xyz))+ 1(yzsin(xyz)+ 2)+ 1(- yzsin(xyz)= 2.

 

[supercali]

thanks dude
i really should sleep more because i simply made a calculation mistake this exam in calculus 2 will drive me nuts
thanks for the help alllot!