## Work to pump out a tank

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### 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?
Last edited by aaron-math on Thu Oct 27, 2011 7:48 am, edited 1 time in total.

aaron-math
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### 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.
galactus
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### Re: Work to pump out a tank

Why did you set the center at (0,5)?

aaron-math
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Posts: 331
Joined: Fri Sep 09, 2011 1:12 pm
Location: New York, NY