# Thread: Double integral, switching to polar coordinates

1. ## Double integral, switching to polar coordinates

$
\iint_{\text{R}} {{\text{exp( - x}}^{\text{2}} {\text{ - y}}^{\text{2}} {\text{)dxdy}}}
$

,where R is the quadrant x>0, y>0

$
\iint_{\text{R}} {{\text{exp( - (x}}^{\text{2}} {\text{ + y}}^{\text{2}} {\text{))dxdy}}}
$

$
\iint_{} {{\text{exp( - r}}^{\text{2}} {\text{)rdrd}}\theta }
$
]

Ok so I've done a couple of these now, but I always seem to have the problem of picturing this geometrically. Can anyone explain a simple way of knowing how equations like these will look on a graph?

what does it mean 'where R is in the quadrant, x> 0, y>0 ??

I always have problems when trying to define the limits for these types of integrals..

2. Originally Posted by dankelly07
$
\iint_{\text{R}} {{\text{exp( - x}}^{\text{2}} {\text{ - y}}^{\text{2}} {\text{)dxdy}}}
$

,where R is the quadrant x>0, y>0

$
\iint_{\text{R}} {{\text{exp( - (x}}^{\text{2}} {\text{ + y}}^{\text{2}} {\text{))dxdy}}}
$

$
\iint_{} {{\text{exp( - r}}^{\text{2}} {\text{)rdrd}}\theta }
$
]

Ok so I've done a couple of these now, but I always seem to have the problem of picturing this geometrically. Can anyone explain a simple way of knowing how equations like these will look on a graph?

what does it mean 'where R is in the quadrant, x> 0, y>0 ??

I always have problems when trying to define the limits for these types of integrals..
The restrictions $x>0,y>0$ imply that you're looking at the first quadrant. In terms of polar coordinates, you're considering a region with radius starting from the pole, and it goes to infinity. Theta goes from 0 to $\frac{\pi}{2}$. So you want to consider evaluating the integral $\int_0^{\frac{\pi}{2}}\int_0^{\infty}e^{-r^2}r\,dr\,d\theta$