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May 18th, 2014, 02:42 PM   #1
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Joined: Apr 2014
From: australia

Posts: 68
Thanks: 32

double integral problem

Hi,
Trying to solve the attached double integral, not sure if I have done it correctly, can someone check? (see attachment).

thanks
Attached Files
 Q3.pdf (19.3 KB, 6 views)

 May 18th, 2014, 06:06 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,618 Thanks: 2608 Math Focus: Mainly analysis and algebra I get almost the same result, you've just made an error in the last step. $$r^2 + 1 \ne 0 \; \text{where r = 0}$$ Thanks from harley05
 May 18th, 2014, 07:13 PM #3 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 thanks for checking, could please show me your working at the point where i have made a mistake. thanks for your time.
 May 18th, 2014, 07:25 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,618 Thanks: 2608 Math Focus: Mainly analysis and algebra I worked the integral slightly differently, but your error isn't in the integral. You have $$I = \left. \frac{\pi}{4}\left[ \left(r^2 + 1\right) \ln{\left( r^2 + 1 \right)} - \left( r^2 + 1 \right) \right] \right|_0^3$$ Which I agree with. But the next line (the last one is wrong). I'm reasonably sure that you evaluated the expression at $r = 3$ correctly, but I think you got 0 at $r = 0$, which is not correct. Thanks from harley05
 May 18th, 2014, 07:35 PM #5 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 yes you're right, thanks for your help. I get 11.02 now. Last edited by harley05; May 18th, 2014 at 07:39 PM.
 May 18th, 2014, 07:51 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,618 Thanks: 2608 Math Focus: Mainly analysis and algebra Me too. Thanks from harley05

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