# Thread: Work Problem: Hemispherical Tank

1. ## Work Problem: Hemispherical Tank

A tank is a hemisphere of radius 12 feet with flat base on the ground and is filled to a height of 9 feet with liquid nitrogen which has a density of 50.45lb/ft^3. How much work is needed to pump all of the hydrogen to the top of the tank?

MY WORK:

Work = volume*density*distance

The equation of the circle is x^2 +y^2 = r^2
Given radius = 12, therefore x^2+y^2=144; x^2=144-y^2

Force = volume * density

Density of nitrogen is 50.45 lbs/ft^3.
Volume = pi*x^2 dy;

Force = [pi*x^2 dy]*[50.45]
Force = 50.45pi[x^2 dy]

Distance = (-y) because all y-coordinates below origin is negative.

So,
Work = [(50.45pi)(x^2)dy]*(-y)

Plugging in 144-y^2 for x^2 by equation of circle:

Work = [(50.45pi)(144 - y^2)dy]*(-y)
Work = (50.45pi)[y^3 - 144y]dy

Nitrogen = 9 feet deep, therefore integrate by dy from y = -12 to y = -3.

Work = (50.45pi)INT.(-12, -3)[y^3 -144y]dy
Work = (50.45pi)[y^4/4 - 72y^2]|(-12, -3)
Work = (50.45pi)[{(1/4)(-3)^4 - 72(-3)^2} - {(1/4)(-12)^4 - 72(-12)^2}]
Work = (50.45pi)[{81/4 - 648} - {5184 - 10368}]
Work = (50.45pi)[-2511/4 + 5184]
Work = (50.45pi)[4556.25]
Work = APPROX 722135.3231 ft-lbs.

2. I don't think your formulas are correct for force and work:

$W=Fs$

Where:
W = work
F = force
s = distance force is applied over

$F=ma$

Where:
m = mass
a = acceleration

$m=\rho V$

Where:

$\rho$ = density
V = Volume

Putting this together:

$W=\rho Vas$

Where the acceleration will be $9.8ms^{-2}$ needed to overcome gravity.