# Thread: different approach on calculating flux threw area..

1. ## different approach on calculating flux threw area..

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. first of all z = x^2 +y^2 is a parabaloid

secondly 0 < x < 1 and 0 < y < 1

is a square not a circle so you want to stay with rectangular coordinates

with limits 0 to 1 for both x and y

3. thanks

4. what if i have only this area
z^2=16-x^2 -y^2
with out any limitations
here i can use polar coordinated
i put z=0 -> r=4
and it generates from a point so in the start r=0
correct?

5. If z > 0 is a stated condition then In this case you're region in the x-y plane is indeed a circle of radius 4 so convert to polar and r varies from 0 to 4 and theta from 0 to 2pi.