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Kendra N's question at Yahoo! Answers regarding work done to empty a conical tank

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Feb 24, 2012
Here is the question:

Applying integration to physics and engineering?

A tank has the shape of an inverted circular cone with height 10 m and base radius 4 m. It is filled with water to a height of 8 m. Find the work required to empty the tank by pumping all of the water to the top of the tank (the density of water is 1000 kg per cubic meter).

PLEASE HELP! I do not understand how it is explained in the textbook.
Here is a link to the question:

Applying integration to physics and engineering? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
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Feb 24, 2012
Hello Kendra N,

I prefer to work problems like this in general terms, and derive a formula we may then plug our data into. Let's let the base radius of the conical tank be $r$, and the height be $h$.

Let's orient a vertical $y$-axis along the axis of symmetry of the tank, where the origin is at the bottom of the tank, and the initial depth of the fluid in the tank is $y_0$. We wish to find the amount of work $W$ is needed to pump all of the fluid to the top of the tank.

Now, if we imagine slicing the cone of water we wish to remove into disks, we may state, using work is force $F$ times distance $d$ for a constant force:

\(\displaystyle dW=Fd=mgd\)

The force exerted is equal to the weight of the slice, which is the product of the mass and the acceleration due to gravity.

The mass $m$ of the slice is the product of the mass density $\rho$ (in \(\displaystyle \frac{\text{kg}}{\text{m}^3}\)) and the volume \(\displaystyle V=\pi r_s^2\,dy\) of the slice, i.e.:

$m=\rho\pi r_s^2\,dy$

The radius of the slice can be found by similarity. Please refer to the following diagram:


\(\displaystyle \frac{r_s}{y}=\frac{r}{h}\,\therefore\,r_s=\frac{r}{h}y\)

The distance the slice must be vertically moved against gravity is:


Putting it all together, we have:

\(\displaystyle dW=\frac{gr^2\rho\pi}{h^2}\left(hy^2-y^3 \right)\,dy\)

Summing up all the work elements through integration, we obtain:

\(\displaystyle W=\frac{gr^2\rho\pi}{h^2}\int_0^{y_0} hy^2-y^3\,dy\)

Applying the anti-derivative form of the FTOC, there results:

\(\displaystyle W=\frac{gr^2\rho\pi}{12h^2}\left[4hy^3-3y^4 \right]_0^{y_0}=\frac{gr^2\rho y_0^3\pi}{12h^2}\left(4h-3y_0 \right)\)

We now have a formula into which we may plug the given and known data:

\(\displaystyle g=9.8\,\frac{\text{m}}{\text{s}^2}\)

\(\displaystyle r=4\text{ m}\)

\(\displaystyle \rho=1000\,\frac{\text{kg}}{\text{m}^3}\)

\(\displaystyle h=10\text{ m}\)

\(\displaystyle y_0=8\text{ m}\)

and so we find:

\(\displaystyle W=\frac{\left(9.8\,\frac{\text{m}}{\text{s}^2} \right)\left(4\text{ m} \right)^2\left(1000\,\frac{\text{kg}}{\text{m}^3} \right)\left(8\text{ m} \right)^3\pi}{12\left(10\text{ m} \right)^2}\left(4\left(10\text{ m} \right)-3\left(8\text{ m} \right) \right)=\frac{3211264\pi}{3}\,\text{J}\)

To Kendra N and any other guests viewing this topic, I invite and encourage you to post other calculus based physics problems in our Advanced Applied Mathematics forum.

Best Regards,