# Calculate this integral using Stokes

1. Aug 21, 2008

### supercali

1. The problem statement, all variables and given/known data
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 $$(x - 1)^2 + (y - 2)^2 = 4$$ and the plane y+z=3/2
calculate:
$$$\left| {\int {\vec Fd\vec r} } \right|$$$

2. Relevant 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 wont work this that sin...

Last edited by a moderator: Aug 21, 2008
2. Aug 21, 2008

### NoMoreExams

Re: stokes

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

3. Aug 21, 2008

### HallsofIvy

Staff Emeritus
Re: stokes

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

4. Aug 21, 2008

### NoMoreExams

5. Aug 21, 2008

### supercali

Re: stokes

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 dont get it how does it disapear the sin(xyz)????

6. Aug 21, 2008

### HallsofIvy

Staff Emeritus
Re: stokes

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.

Last edited: Aug 21, 2008
7. Aug 21, 2008

### supercali

Re: stokes

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!!!