stocks question..

**transgalactic** · August 13th 2009, 11:20 AM

calculate
$ \iint_{M}^{}rot(\vec{F})\vec{dS} $
where
$ \vec{F}=(y^2z,zx,x^2z^2) $
M is a part of $z=x^2+y^2$ which is located in $1=x^2+y^2$
and its normal vector points outside

i am used to solve it like this
$ \iint_{M}^{}rot\vec{F}\vec{dS}=\iint_{D}\frac{rot\ vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy $
$ \vec{N}=(2x,2y,-1) $
$ \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy $
now i convert into polar coordinates

x^2+y^2=r
is this method ok?

**luobo** · August 13th 2009, 03:12 PM

Originally Posted by transgalactic

calculate
$ \iint_{M}^{}rot(\vec{F})\vec{dS} $
where
$ \vec{F}=(y^2z,zx,x^2z^2) $
M is a part of $z=x^2+y^2$ which is located in $1=x^2+y^2$
and its normal vector points outside

i am used to solve it like this
$ \iint_{M}^{}rot\vec{F}\vec{dS}=\iint_{D}\frac{rot\ vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy $
$ \vec{N}=(2x,2y,-1) $
$ \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy $
now i convert into polar coordinates

x^2+y^2=r
is this method ok?

Stocks question or Stokes question?

**transgalactic** · August 13th 2009, 09:35 PM

i dont know if it stocks or stokes
in general it about integration threw some closed circle etc..

is my method correct

**transgalactic** · August 14th 2009, 01:14 AM

the formal solution is differs so muuch
http://i30.tinypic.com/14cede9.gif

is my method ok?
i didnt solve it with parameters

**transgalactic** · August 14th 2009, 01:48 AM

why are they switching to parameters?

**luobo** · August 14th 2009, 04:19 AM

Originally Posted by transgalactic

why are they switching to parameters?

Switching to parameters because it has the following advantages:

(1) The problem may require you to use the Stoke's Theorem, as you have indicated in the title "stocks question";

(2) Use the Stoke's Theorem reduces the problem to 1D;

(3) Solving the problem in 2D is very complicated, sometimes even impossible.

**transgalactic** · August 14th 2009, 04:22 AM

look at my final integral
its not hard at all

what so hard integrating by x
and then by y

??

**luobo** · August 14th 2009, 04:22 AM

Originally Posted by transgalactic

the formal solution is differs so muuch
http://i30.tinypic.com/14cede9.gif

is my method ok?
i didnt solve it with parameters

You are trying to solve the problem in 2D. That's ok as long as you get everything right and you can work it out. Did you check if you computed the curl $\vec{F}$ right?

curl $\vec{F} = \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right) e_1 - \left(\frac{\partial F_3}{\partial x}- \frac{\partial F_1}{\partial z}\right) e_2 + \left(\frac{\partial F_2}{\partial x}- \frac{\partial F_1}{\partial y}\right) e_3$ .

**transgalactic** · August 14th 2009, 04:27 AM

i would like to solve it by parameters to
but if i will put trigonometric variables in my final integral

its a double integral how it could become a single integral
even if i substitute dx and dy and the rest
its still remains a double integral

**luobo** · August 14th 2009, 04:45 AM

Originally Posted by transgalactic

i would like to solve it by parameters to
but if i will put trigonometric variables in my final integral

its a double integral how it could become a single integral
even if i substitute dx and dy and the rest
its still remains a double integral

The Stokes Theorem reduces the problem from an integral over an surface ( $x^2+y^2= z$ , which is a surface) to an integral along a curve ( $x^2+y^2= 1$ , which is a curve).

**luobo** · August 14th 2009, 08:21 AM

Originally Posted by transgalactic

calculate
$ \iint_{M}^{}rot(\vec{F})\vec{dS} $
where
$ \vec{F}=(y^2z,zx,x^2z^2) $
M is a part of $z=x^2+y^2$ which is located in $1=x^2+y^2$
and its normal vector points outside

i am used to solve it like this
$ \iint_{M}^{}curl\vec{F}\vec{dS}=\iint_{D}\frac{rot \vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy $
$ \vec{N}=(2x,2y,-1) $
$ \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(-x,2xz^2-y^2,z-yz) \cdot (2x,2y,-1)}{1}dxdy $
now i convert into polar coordinates

x^2+y^2=r
is this method ok?

OK. Let's try direct surface integration step by step, instead of using the Stoke's theorem and see how more complicated it is.

$\vec{F}=(y^2z, xz, x^2y^2)$
(Note: the image you provided indicate the third component is $x^2y^2$ , rather than $x^2z^2$ . Anyway, it does not influence the final answer)

$rot \vec{F} = (2x^2y-x)e_1 + (y^2-2xy^2)e_2 + (z-2yz)e_3=$ $(2x^2y-x)e_1+ (y^2-2xy^2)e_2 + ( x^2 + y^2-2x^2y-2y^3)e_3$

$\vec{dS} = (2xe_1 + 2ye_2 -1e_3) dxdy$

$curl \vec{F} \cdot \vec{dS} = -3x^2 - y^2 +2x^2y+4y^3+4x^3y-4xy^3$

$\iint_{M}^{}curl\vec{F} \cdot \vec{dS}=\iint_{x^2+y^2\leq 1} ($ $-3x^2 - y^2 +2x^2y+4y^3+4x^3y-4xy^3)\, dxdy$

Taking advantage of the center symmetry,
$\iint_{M}^{}curl\vec{F} \cdot \vec{dS}=\iint_{x^2+y^2\leq 1} ($ $-3x^2 - y^2)\, dxdy$

Let $x=r\cos\theta, y=r\sin\theta$
$\iint_{M}^{}curl\vec{F} \cdot \vec{dS}=$ $-\int_{0}^{2\pi}\int_{0}^{1} (3\cos^2\theta+\sin^2\theta)\,r^3 \, dr d\theta$ $=-\frac{1}{4}\int_{0}^{2\pi} (3\cos^2\theta+\sin^2\theta)\, d\theta=-\frac{1}{4}\int_{0}^{2\pi} (2+\cos 2\theta)\, d\theta=-\pi$

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