Calculus II Work and Density Question

[ardentmed]

Hey guys,

View attachment 2808
I'm not too sure of my answer for this question.

Knowing the the area is πr^2 for a single slice, I setup an integral with radius =5 and got $ \int_{0.3}^{1.5} 245,000πx \,d $ ultimately leads to 264,000π Joules for the answer.

Of course, this is bearing in mind that F=mg and D=m/v
Am I on the right track?

Thanks in advance.

 

[MarkFL]

Let's develop a general formula into which we can just plug the given data:

Suppose we have an upright cylindrical tank, and we wish to compute the amount of work required to pump some of the fluid out, to some distance above the top of the tank.

Let's let the radius of the cylindrical tank be $r$, the height of the tank be $h$ and the additional distance above the tank the fluid must be pumped be $q$. All measures are in meters.

Let's then imagine the surface of the fluid is initially a distance $h_1$ from the top of the tank, and we wish to pump out fluid until the surface is a distance $h_2$ from the top of the tank. We will let $W$ be the amount of work required to accomplish this task.

We will use a vertical $y$-axis along which to integrate. Now, if we imagine slicing the cylinder of water we wish to remove into disks, we may state, using work is force times distance 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 \(\displaystyle V=\pi r^2\,dy\) of the slice, i.e.:

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

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

$d=q+y$

Putting it all together, we have:

\(\displaystyle dW=\rho\pi g r^2(q+y)\,dy\)

Now, to find the total work done, we may sum the work differentials through integration:

\(\displaystyle W=\rho\pi g r^2\int_{h_1}^{h_2}(q+y)\,dy\)

Applying the anti-derivative form of the FTOC, we find:

\(\displaystyle W=\rho\pi g r^2\left[qy+\frac{1}{2}y^2 \right]_{h_1}^{h_2}\)

\(\displaystyle W=\rho\pi g r^2\left(\left(qh_2+\frac{1}{2}h_2^2 \right)-\left(qh_1+\frac{1}{2}h_1^2 \right) \right)\)

\(\displaystyle W=\rho\pi g r^2\left(q(h_2-h_1)+\frac{1}{2}(h_2^2-h_1^2) \right)\)

\(\displaystyle W=\rho\pi g r^2(h_2-h_1)\left(q+\frac{1}{2}(h_2+h_1) \right)\)

\(\displaystyle W=\frac{1}{2}\rho\pi g r^2(h_2-h_1)\left(h_1+h_2+2q \right)\)

Okay, now we have a general formula. Let's identify the parameters from your given problem.

\(\displaystyle \rho=\frac{1000\text{ kg}}{\text{m}^3},\,g=9.8\frac{\text{m}}{\text{s}^2}\,r=5\text{ m},\,h_1=0.3\text{ m},\,h_2=1.5\text{ m},\,q=0\text{ m}\)

And so we find:

\(\displaystyle W=\frac{\pi}{2}\left(\frac{1000\text{ kg}}{\text{m}^3}\right)\pi\left(9.8\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)^2\left(1.5\text{ m}-0.3\text{ m}\right)\left(0.3\text{ m}+1.5\text{ m}+2\cdot0\text{ m} \right)\)

\(\displaystyle W=264600\pi\text{ J}\)

 

[ardentmed]

Excellent. Thank you so much for the insightful response.

But wait, what accounts for the extra 400π between our answers? Our setups for the question seems relatively similar.

 

[MarkFL]

Excellent. Thank you so much for the insightful response.

But wait, what accounts for the extra 400π between our answers? Our setups for the question seems relatively similar.

You've simply made an arithmetical error. When I evaluate the integral you set up, I get the same result I posted above. :D