# Different approach on calculating flux threw area

1. Aug 13, 2009

### slonopotam

calculate
$$\iint_{M}^{}\vec{F}\vec{dS}$$
where
$$\vec{F}=(e^y,ye^x,x^2y)$$
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
$$\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{\vec{F}\cdot \vec{N}}{|\vec{N}\cdot\vec{K}|}dxdy$$
$$\vec{N}=(2x,2y,-1)$$
$$\iint_{M}^{}\vec{F}\vec{dS}=\iint_{D}\frac{(e^y,ye^x,x^2y) \cdot (2x,2y,-1)}{1}dxdy$$
now i convert into polar coordinates

x^2+y^2=r
$$=\int_{0}^{2\pi}\int_{0}^{1}\frac{(e^y,ye^x,x^2y) \cdot (2x,2y,-1)}{1}rdrd\theta$$
how to what are the intervals for r
i just guessed its from 0 to 1
i dont know how to know the upper interval here

except that
is this method ok?

2. Aug 13, 2009

### gabbagabbahey

$x^2+y^2$ 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 $\sqrt{1^2+1^2}=\sqrt{2}$?

3. Aug 13, 2009

### kuruman

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?

4. Aug 13, 2009

### gabbagabbahey

Nope, my mistake...just a small brainfart