My Math Forum Work to pump out a tank

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 October 25th, 2011, 09:04 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Work to pump out a tank The hemispherical tank shown is full of water. Given that water weighs $62.5 lb/ft^3$, find the work required to pump the water out of the tank. My Attempt: The volume of a sample cross section is $v= \pi r^2 \,dx$ $v=25 \pi dx$ The mass of the cross section is $mass= (volume)(density)$ $v=1562.5 \pi dx$ The force need to lift this section is $F=mg$ $v=15312.5 \pi \, dx$ The work needed to lift this section is $w=Fd$ $v=15312.5 \pi x \, dx$ The do not give limits so I am assuming [a b] Integrating we have $15312.5 \pi \int_a \,^b x \, dx$ $15312.5 \pi [ \dfrac{x^2}{2}]_a ^b$ finally giving $7656 \pi [ a^2-b^2]$ I don't know if this is correct?
 October 26th, 2011, 11:49 AM #2 Senior Member   Joined: May 2011 Posts: 500 Thanks: 4 Re: Work to pump out a tank For a circle of radius 5 with center (0,5), we have $x^{2}+(y-5)^{2}=25$ $x^{2}=10y-y^{2}$ The force increment can be written as: $62.5(\pi x^{2} dy)=62.5 \pi (10y-y^{2})dy$ The disc y feet from the bottom must be moved a distance of 5-y feet. So, we have $62.5 \pi (10y-y^{2})(5-y)=62.5\pi (y^{3}-15y^{2}+50y)$ $62.5\pi \int_{0}^{5}(y^{3}-15y^{2}+50y)dy$ Another way to set it up: $62.5\pi \int_{-5}^{0}(25-y^{2})(-y)dy$ Both give the same result.
 October 27th, 2011, 06:15 AM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Work to pump out a tank Hey, Thx for the reply/help. Why did you set the center at (0,5)?

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