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March 2, 2015

March 2, 2015

Posted by **Alex** on Thursday, March 22, 2012 at 5:44pm.

- Integrals/Related Rates(?)/Volume -
**bobpursley**, Thursday, March 22, 2012 at 6:09pmok, to get the flow, you integrate area*flow rate thru that area.

so dArea= dr*2PI

dVolume= flowrate*dArea

Volume= INT 9r^2*2PI*dr=18PI r^3/3 from r=0 to r=3

volumeflowrate= 6PI *27 cm^3/sec

total amount H2O: volumeflowrate*time

- Integrals/Related Rates(?)/Volume -
**Damon**, Thursday, March 22, 2012 at 6:11pmthrough every ring of radius r and thickness dr and amount per second flows equal to 2 pi r v dr

or

dq = 2 pi (9 r - r^3) dr

integrate that from 0 to 3 and then multiply by 4 seconds

18 pi (r^2/2) - 2 pi (r^4/4)

9 pi (9) - (pi/2)(81)

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