http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Mon, 21 Jul 2014 11:01:34 -0500http://calc3.askbot.com/question/147/improper-double-integrals/In section 15.2, question #19a, it asks: Consider the integral $\int\int 1/\sqrt{x^2+y^2} dA$, where $D$ is the unit disk centered at the origin. a) Why might this integral be considered improper? The link to see this graph did not work for me, but I used wolfram alpha and the graph looked kind of like a piece of paper being folded upwards. I don't know why this could be considered improper? Any thoughts?Mon, 21 Jul 2014 10:17:19 -0500http://calc3.askbot.com/question/147/improper-double-integrals/http://calc3.askbot.com/question/147/improper-double-integrals/?answer=148#post-id-148To talk about this we need to remember what makes an integral improper. An integral is improper when it is bound by an asymptote or infinity. When $x$ and $y$ are equal to zero in this function the function will be undefined and $z$ values will become arbitrarily large as $x$ and $y$ approach zero. So your integral bounds for the $dr$ part of the problem will be at an asymptote. This makes the integral improper.Mon, 21 Jul 2014 11:01:34 -0500http://calc3.askbot.com/question/147/improper-double-integrals/?answer=148#post-id-148