Different approach on calculating flux threw area

[slonopotam]

calculate
<br /> \iint_{M}^{}\vec{F}\vec{dS}<br />
where
<br /> \vec{F}=(e^y,ye^x,x^2y)<br />
M is a part of hyperboloid x^2+y^2
which is located at 0<=x<=1 and 0<=y<=1 ,and its normal vector points outside
like this :
http://i28.tinypic.com/f9p63r.gif

i am used to solve it like this
<br /> \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{\vec{F}\cdot<br /> \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy<br />
<br /> \vec{N}=(2x,2y,-1)<br />
<br /> \iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(e^y,ye^x,x^2y) \cdot<br /> (2x,2y,-1)}{1}dxdy<br />
now i convert into polar coordinates

x^2+y^2=r
<br /> =\int_{0}^{2\pi}\int_{0}^{1}\frac{(e^y,ye^x,x^2y) \cdot (2x,2y,-1)}{1}rdrd\theta<br />
how to what are the intervals for r
i just guessed its from 0 to 1
i don't know how to know the upper interval here

except that
is this method ok?

 

[gabbagabbahey]

M is a part of hyperboloid x^2+y^2

x^2+y^2 is just a meaningless expression and does not describe a hyperboloid. Do you mean z^2=x^2+y^2?

now i convert into polar coordinates

x^2+y^2=r

You mean x^2+y^2=r^2, right?

how to what are the intervals for r
i just guessed its from 0 to 1
i don't know how to know the upper interval here

Well, x^2+y^2=r^2 and x and y each go from 0 to 1, so wouldn't r go from \sqrt{0^2+0^2}=0 to \sqrt{1^2+1^2}=\sqrt{2}?

 

[kuruman]

x^2+y^2 is just a meaningless expression and does not describe a hyperboloid. Do you mean z^2=x^2+y^2?



You mean x^2+y^2=r^2, right?



Well, x^2+y^2=r^2 and x and y each go from 0 to 1, so wouldn't r go from \sqrt{0^2+0^2}=0 to \sqrt{1^2+1^2}=\sqrt{2}?

I am not sure about the square root of 2; Since at constant z you have a circle in a plane parallel to xy, x and y are related to r by the parametric equations

x = r cosφ y = r sinφ

When x = 1, y = 0. Am I missing something?

 

[gabbagabbahey]

When x = 1, y = 0. Am I missing something?

Nope, my mistake...just a small brainfart