Welcome to our community

Be a part of something great, join today!

Hazue's question at Yahoo! Answers: work required to pump water from a hemispherical tank

  • Thread starter
  • Admin
  • #1


Staff member
Feb 24, 2012
Here is the question:

Hemispherical dome - Calculus II question?

I have a problem that I can't figure out how to set up. I'm really bad at word problems.

1) A hemispherical dome of radius 10 meters is filled with water to a depth of 6 meters. How much work is needed to pump out water through the top of the dome until the depth is 2 meters? (Note: the dome's flat side rests on flat ground.)

An explanation would be really helpful. The set up is the part that I struggle with the most, so if anyone could provide it, that would be helpful.

I have posted a link there to this thread so the OP can view my work.
  • Thread starter
  • Admin
  • #2


Staff member
Feb 24, 2012
Hello Hazue,

I prefer to work problems like this in general terms, and then plug our given data into the resulting formula.

First, let's let:

$R$ = the radius of the hemispherical tank.

$\rho$ = the weight density of the fluid.

$g$ = the acceleration due to gravity.

Now, let's imagine slicing the contents of the tank horizontally into circular sheets. The radius of each sheet will be a function of its distance from the base of the tank. So, let's orient a vertical $y$-axis passing through the axis of symmetry of the tank, with its origin at the base of the tank.

The circular portion of the cross-section of the tank containing our $y$-axis will lie on the circle:

\(\displaystyle x^2+y^2=R^2\)

And so, for any circular sheet, its radius will be the $x$-coordinate of a horizontal line passing through the circle, and so we may state:

\(\displaystyle r^2=R^2-y^2\)

Hence, the volume of an arbitrary sheet can be given by:

\(\displaystyle dV=\pi\left(R^2-y^2 \right)\,dy\)

Next, we want to determine the weight $w$ of this sheet. Using the definition of weight density, we may state:

\(\displaystyle \rho=\frac{w}{dV}\,\therefore\,w=\rho\,dV\)

Next, we observe that the distance $d$ this sheet must be lifted is:

\(\displaystyle d=R-y\)

Thus, using the fact that work is the product of the applied force and the distance through which this force is applied, we find the work done to lift the arbitrary sheet is:

\(\displaystyle dW=wd=\pi\rho(R-y)\left(R^2-y^2 \right)\,dy\)

Distributing, we find:

\(\displaystyle dW=\pi\rho\left(R^3-R^2y-Ry^2+y^3 \right)\,dy\)

Now, if $y_i$ is the initial depth of fluid in the tank, and $y_f$ is the final depth, where $0\le y_f<y_i\le R$, then the total amount of work required to pump out the required amount of fluid is given by:

\(\displaystyle W=\pi\rho\int_{y_f}^{y_i} R^3-R^2y-Ry^2+y^3\,dy\)

Applying the FTOC, there results:

\(\displaystyle W=\pi\rho\left[R^3y-\frac{1}{2}R^2y^2-\frac{1}{3}Ry^3+\frac{1}{4}y^4 \right]_{y_f}^{y_i}=\frac{\pi\rho}{12}\left[12R^3y-6R^2y^2-4Ry^3+3y^4 \right]_{y_f}^{y_i}\)

And so we find:

\(\displaystyle W=\frac{\pi\rho}{12}\left(y_i-y_f \right)\left(12R^3-6R^2\left(y_i-y_f \right)-4R\left(y_i-y_f \right)^2+3\left(y_i-y_f \right)^3 \right)\)

Now, we are given the following data for this particular problem:

\(\displaystyle \rho=\frac{1000\text{ kg}}{\text{m}^3}\cdot9.8\frac{\text{m}}{\text{s}^2}=9800\frac{\text{N}}{\text{m}^3},\,R=10\text{ m},\,y_i=6\text{ m},\,y_f=2\text{ m}\)

Plugging these values into our formula, we obtain:

\(\displaystyle W=\frac{2450\pi}{3}(6-2)\left(12(10)^3-6(10)^2(6-2)-4(10)(6-2)^2+3(6-2)^3 \right)\text{ J}=\frac{89689600\pi}{3}\text{ J}\)