# Problem

A hemispherical tank of radius 2 feet is positioned so that its base is circular. How much work is required to fill the tank with water through a hole in the base when the water source is at the base? (The weight-density of water is 62.4 pounds per cubic foot.)

To find the Work I first needed to find the Force because Work = force*distance. To do this I am using the "disc-method" which subdivides a 3d solid into discs, in this case, a dome.

## Finding the force

To find the force I first need to find the volume of each disc and then multiply by the density, which the problem says is 62.4 lb/ft, so I only need to find the volume of each disc.

The Volume of a disc is $$\pi r^2h$$ so while the height of each disc will end up being equivalent to dy, I will have to find radius as a function of y for every disc.

###### Radius of each disc

Attempting to follow this video, I was able to see that I could write the radius of each disc as a function of y by using the Pythagorean theorem.

In this case, because $$x^2+y^2=z^2$$ and $$z^2$$ can be equal to the radius, that gives $$x^2+y^2=2$$. Finally, simplifying, showing x as a function of y:

$$x^2=2-y^2$$

Now that I have the radius in terms of y, I can plug that back into the formula for the volume of our discs giving: $$\pi (2-y^2)dy$$

Then, multiplying by the given density of water(62.4lb/ft), we get the force: $$\pi (2-y^2)62.4dy$$

Then for every disc, and not just some disc: $$\int_0^2 {\pi (2-y^2)62.4}dy$$

### Finishing up

Now that I have the equation for the force of every disc combined, I just need to multiply by the distance traveled by each disc which is the same as the height of the dome(2ft.), which gives:

$$\int_0^2 {\pi (2-y^2)62.4*2}dy = 522.761$$

Unfortunately though, this is incorrect, and I'm not sure where I've gone wrong.

• Each disk traveled different distances, otherwise all your water float at elevation 2ft Commented Sep 20, 2021 at 10:52
• @RezhaAdrianTanuharja would $2-y$ properly represent the distance traveled by each?
– Papu
Commented Sep 20, 2021 at 11:30
• You actually have not just one error. You have at least two. Commented Sep 20, 2021 at 12:19
• $x^2 + y^2 = 2^2$ Commented Sep 21, 2021 at 2:32

Ignoring friction etc., it's just like pumping up the same amount of water to a height where the center of mass of the hemisphere is located. For the center of mass of a hemisphere, see this question.. Hence the needed energy is $$E = \rho\cdot V \cdot g\cdot h_\text{center_of_mass}$$

• $$\rho$$ is the density of the fluid
• $$V$$ is its volume ($$2\pi/3r^3)$$
• $$g$$ is gravitational acceleration (assuming it is constant, $$\approx9.81\mathrm{m/s^2}$$)

If you are imperial, be aware of conversion factors.

Note: One error is that you are using $$x^2=r-y^2$$ where the correct one is $$x^2 = r^2-y^2$$ with $$r^2=4$$.

Second error is that you are integrating the weights (forces) of the slices, which gives a force and this cannot be right, because the outcome will have dimension of force, but should be energy. Gravitational potential energy is Force × Displacement (provided force and displacement are parallel, which is the case here). Thus the integrand is missing a factor of $$y$$.

Hence a final recommendation: Resist the temptation to plug in constants and then have to carry around magic values (and their units). Try to derive a formula, and then finally plug in all the values and check that the dimensionality works out correctly like kg·m2/s2 or poundmass·foot2/s2

• @Papu: ...and if you really want to use or must use integration, you can at least use this one to check your outcome :-) Commented Sep 20, 2021 at 12:00