Elizabeth's question at Yahoo Answers regarding the computation of work

[MarkFL]

Here is the question:

Calculus help upper and lower half?

A cylinder with height 8 m and radius 3 m is filled with water and must be emptied through an outlet pipe 2 m above the top of the cylinder.

(A) compute the work required to empty the water in the top half of the tank
(B) compute the work required to empty the equal amount of water in the lower half of the tank
(C) interpret

Please show any steps you used. Thanks!

Here is a link to the question:

Calculus help upper and lower half? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Elizabeth,

I prefer to work problems in general terms, and derive a formula we may then plug our data into. 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:

$$dW=Fd=mgd$$

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 $$\frac{\text{kg}}{\text{m}^3}$$) and the volume $$V=\pi r^2\,dy$$ 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:

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

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

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

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

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

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

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

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

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

Now we may plug in the given/known data

Water is generally taken to have a mass density of:

$$\rho=1000\,\frac{\text{kg}}{\text{m}^3}$$

I will assume we are to use:

$$g=9.8\,\frac{\text{m}}{\text{s^2}}$$

We are given:

$r=3\text{ m},\,q=2\text{ m},\,h=8\text{ m}$

For part (A), we have:

$$h_1=0\text{ m},\,h_2=\frac{h}{2}=4\text{ m}$$

Hence:

$$W_A=1411200\pi\text{ J}$$

For part (B), we have:

$$h_1=\frac{h}{2}\,\text{m}=4\text{ m},\,h_2=h=8\text{ m}$$

Hence:

$$W_B=2822400\pi\text{ J}=2W_A$$

(C) If we interpret the amount of work done as the area of a trapezoid, since the integrand is a linear function, we find that:

$$W_A=k\cdot\frac{1}{2}(6+2)4=16k$$

$$W_B=k\cdot\frac{1}{2}(10+6)4=32k=2W_A$$