asked 2014-07-21 10:17:19 -0600
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?
answered 2014-07-21 11:01:34 -0600
To 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.
Asked: 2014-07-21 10:17:19 -0600
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Last updated: Jul 21 '14
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