I'm not sure what this question is asking for. If someone could explain, that would be wonderful. Pictures or diagrams would greatly be appreciated, but as I said, I'm just not sure what the question is asking for.

Background Info:
Oil flows through a cylindrical pipe with a radius of 3 in, but friction slows the flow towards the outer edge. The speed at which the oil flows at a distance r inches from the center is 8(10-r^2) inches/sec.

Question:
In a plane cross section of the pipe, a thin ring with a thickness Δr at a distance r inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip and hence the approximate area of the ring?

Any help would be appreciated.

I'm quite confused... Is there any other info or question that goes along with it?

There's another question that says:
Explain why we know that oil passes through this ring at approximately 8(10-r^2)2(pi)rΔr cubic inches per second.

But that doesn't help me either. Ah, my dilemmas ;_;

Ok, my guess here:

Say you are looking at the cross-section of the pipe from above. You'll see a circle. In that circle, we have a ring, with thickness \Delta r, after a distance of r from the centre. That makes three circles in all (1 for pipe, 2 for inner ring)

You can find the area of the smallest circle. You can also find the area of the medium circle. (in terms of r and delta r). Subtract the medium area by the smallest area to have the area of the strip.

I guess it means 'approximately' rectangular strip because the surface is initially circular, hence sort of confirming my guess.

Now, I'm not sure whether the 'r' distance is the distance from the centre to the inner or outer edge of the ring. However, it's more likely to be to the inner ring. So, the area will be

Area = \pi (r+\Delta r)^2 - \pi r^2

[In either case, the area happen to become the same]

And the 'hence' part is the same answer as the area of the strip.

Now, I'll think of the other part. Seems to be integration...

EDIT: I accidentally used the formula for circumference rather than area.

Arg... that doesn't seem to make sense to me... the numbers aren't 'falling in place'.

From my reasoning, the oil flows fast at the centre (80 in/s) and slower at the edge of the ring (8(10-r^2) in/s). To find the volume passing through it, you have to consider all the different speeds between the centre and the edge of the ring r. That is found usually by integration... or I'm missing something :(

It is integration (at least it should be). I'll try to solve it on my own.

Thanks big time. I'll just see what I get.

I'm not sure what this question is asking for. If someone could explain, that would be wonderful. Pictures or diagrams would greatly be appreciated, but as I said, I'm just not sure what the question is asking for.

Background Info:
Oil flows through a cylindrical pipe with a radius of 3 in, but friction slows the flow towards the outer edge. The speed at which the oil flows at a distance r inches from the center is 8(10-r^2) inches/sec.

Question:
In a plane cross section of the pipe, a thin ring with a thickness Δr at a distance r inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip and hence the approximate area of the ring?

Any help would be appreciated.




This is definitely an application in integration problem.

Since the ring can be represented by a rectangle, it has area:

V=2{\pi}r\cdot {\Delta}r

You know that the circumference of a circle is 2{\pi}r.

That is equal to the length of the rectangle.

The thickness is the {\Delta}r.

Therefore, the area of the pipe cross section is

2{\pi}\int_{0}^{3}rdr=9{\pi}

Which we can easily get just by using the area of a circle formula.

{\pi}(3)^{2}=9{\pi}

It would appear you may be trying to find the flow rate of the pipe?

By counting up the area of all these rings, we find the area of the pipe cross-section.

Then, since it is velocity times area, multiply this by the flow rate and we have cubic inches per second.

(2{\pi}r{\Delta}r) in^{2}\cdot 8(10-r^{2})\frac{in}{sec}

See? in^{2}\cdot \frac{in}{sec}=\frac{in^{3}}{sec}

Area times velocity gives units of volume/time. The speed is high toward the center and slower at the outer edge. Check it and see.

In the formula, 8(10-r^2), plug in r=1 and r=3 and see which is larger.

The r=1 is because it is closer to the center. It is rather intuitive that the smaller you make a pipe the faster a fluid will shoot through it.

This has a name. It's called the equation of continuty in Fluid Mechanics.


Let's add another part to this problem.

Let's say we want to know the rate at which the water flows through the whole pipe.

16{\pi}\int_{0}^{3}r(10-r^{2})dr=396{\pi} \;\ \frac{in^{3}}{sec}