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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
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The region is called an annulus. Try changing to polar coordinates.
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October 17th, 2018, 01:10 PM   #3
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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 couldn’t connect the on between.
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October 17th, 2018, 02:35 PM   #4
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Let $u=r^2+3$, then $\displaystyle 2\pi\int_2^3\sqrt{r^2+3}rdr = \pi\int_7^{12}\sqrt{u}du=\frac{2}{3}\pi(12\sqrt{12 }-7\sqrt{7})$.
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Last edited by skipjack; October 17th, 2018 at 07:44 PM.
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