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 May 12th, 2014, 02:35 AM #1 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 double intergration hi, really stuck on this, any help would be appreciated. thanks Q3.jpg
 May 12th, 2014, 11:50 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,621 Thanks: 2609 Math Focus: Mainly analysis and algebra On my phone, that picture is virtually unreadable. However, I assume we have $x$s and $y$s which we need to turn into $r$s and $\theta$s. Write $x = r\cos\theta$ and $y = r\sin\theta$ and substituting into the integrand. You then need to change the limits of integration so that the same region is being integrated over. E.g. If the original is integrating $y$ between 0 and $\sqrt{x^2 + y^2}$ and $x$ between 0 and 2, we aare looking at a quarter disc of radius 2. So we integrate with respect to $x$ between 0 and 2 and with respect to $\theta$ between 0 and $\frac{\pi}{2}$.
 May 15th, 2014, 02:27 AM #3 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 i get an answer of; $\displaystyle pi/4(10ln10-10) = 10.23$ not sure if i've did this right? Can anyone please confirm.

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### double intergration

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