October 16th, 2018, 09:06 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 118 Thanks: 6  double integrals
Hello everyone. How can I approach this question? I know that the integral boundaries are 2 to 3 donut shape.. I stuck 
October 16th, 2018, 10:44 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,878 Thanks: 1834 
The region is called an annulus. Try changing to polar coordinates.

October 17th, 2018, 01:10 PM  #3 
Senior Member Joined: Apr 2017 From: New York Posts: 118 Thanks: 6 
I idea convert to polar form and changed the limits for r23 and for theta 02pi And this is what I got. Calculator says 2pi(....) I couldnâ€™t connect the on between. 
October 17th, 2018, 02:35 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622 
Let $u=r^2+3$, then $\displaystyle 2\pi\int_2^3\sqrt{r^2+3}rdr = \pi\int_7^{12}\sqrt{u}du=\frac{2}{3}\pi(12\sqrt{12 }7\sqrt{7})$.
Last edited by skipjack; October 17th, 2018 at 07:44 PM. 

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