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May 12th, 2014, 02:35 AM   #1
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double intergration

hi, really stuck on this, any help would be appreciated.
thanks
Q3.jpg
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May 12th, 2014, 11:50 AM   #2
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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 $\frac{\pi}{2}$.
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May 15th, 2014, 02:27 AM   #3
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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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