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October 16th, 2018, 09:06 PM   #1
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double integrals

Hello everyone.
How can I approach this question?

I know that the integral boundaries are 2 to 3 donut shape..
I stuck
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 October 16th, 2018, 10:44 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,109 Thanks: 1909 The region is called an annulus. Try changing to polar coordinates.
October 17th, 2018, 01:10 PM   #3
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Joined: Apr 2017
From: New York

Posts: 119
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I idea convert to polar form and changed the limits for r2-3 and for theta 0-2pi
And this is what I got.
Calculator says 2pi(....)
I couldnâ€™t connect the on between.
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 October 17th, 2018, 02:35 PM #4 Global Moderator   Joined: May 2007 Posts: 6,665 Thanks: 651 Let $u=r^2+3$, then $\displaystyle 2\pi\int_2^3\sqrt{r^2+3}rdr = \pi\int_7^{12}\sqrt{u}du=\frac{2}{3}\pi(12\sqrt{12 }-7\sqrt{7})$. Thanks from Leonardox Last edited by skipjack; October 17th, 2018 at 07:44 PM.

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