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Math Help - Double integral, switching to polar coordinates

  1. #1
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    Double integral, switching to polar coordinates

    <br />
\iint_{\text{R}} {{\text{exp( - x}}^{\text{2}} {\text{ - y}}^{\text{2}} {\text{)dxdy}}}<br />

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

    <br />
\iint_{\text{R}} {{\text{exp( - (x}}^{\text{2}} {\text{ + y}}^{\text{2}} {\text{))dxdy}}}<br />

    <br />
\iint_{} {{\text{exp( - r}}^{\text{2}} {\text{)rdrd}}\theta }<br />
]


    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..
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dankelly07 View Post
    <br />
\iint_{\text{R}} {{\text{exp( - x}}^{\text{2}} {\text{ - y}}^{\text{2}} {\text{)dxdy}}}<br />

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

    <br />
\iint_{\text{R}} {{\text{exp( - (x}}^{\text{2}} {\text{ + y}}^{\text{2}} {\text{))dxdy}}}<br />

    <br />
\iint_{} {{\text{exp( - r}}^{\text{2}} {\text{)rdrd}}\theta }<br />
]


    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
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