More Calculus: Paraboloid Tank

[Dravin]

Okay I've got a paraboloid tank with a height of 4 feet and a radius of 4 feet that I need to figure out the work to pump the water out of. Some simple math lets us know that the formula for the parabola the tank 'comes' from is y=1/4*x^2.

So, as far as I've been taught to set-up such problems, I figure out the formula for volume based on cylinders. So the generic is going to be v=pir^2*h.. I'm having difficulties but I suspect (rather strongly) my issue is in substituting r. The formula r=2sqrt(4-xi) produces the proper radius. The only problem is when put into the volume formula for the 'disks' being pumped out is that I end up with v=4pi(4-xi). When that get's combined with density of water to get me my weight (we're given the figure of 62.5) when all is said and done I end up with 2000pi which is incorrect (integrating 250pi(4-x) from 0 to 4)*. The correct answer is supposedly 8000pi/3. The /3 suggest to me that my formula needs to have xi (which turns into x) squared before it gets integrated, but it I can't seem to figure out where.

* Online Definite Integral Calculator : %pi*250*(4-x)

[Vort]

Let's make differential-thickness disks and lift them out one by one. As you've noted, the defining parabola's equation is

y = ¼x²

or, solving for x,

x = 2√y

Each disk has radius r = x, and so has area

A = πr²

= πx²

= π(2√y)²

= 4πy

So the differential volume for each disk is

dV = 4πy dy

Since the density ϱ is constant, the weight of each disk is just ϱ dV.

The work to raise each disk is the disk's weight times the distance to the top (y = 4). For any given disk, that height is (4 - y). Thus, for each disk, the differential work is

dW = (4 - y) ϱ dV

= (4 - y) ϱ (4πy dy)

= 4πϱ (4y - y²) dy

Now we can do the integral:

W = ∫ 4πϱ (4y - y²) dy {between y=0 and y=4}

= 4πϱ ∫ (4y - y²) dy {between y=0 and y=4}

= 4πϱ [2y² - y³/3] {between y=0 and y=4}

= 4πϱ [32 - 64/3 - 0 + 0]

= 4πϱ (32/3)

= 128πϱ/3

Given a value of ϱ=62.5 lbs/ft³ (it's actually closer to 62.4, but whatever), that makes the work

128π(62.5)/3 = 8000π/3 ft-lbs

(Note how cleverly I added the units back in here -- they should have been used all along, but I left them off to reduce clutter. But seriously, you should ALWAYS show your units all the way along.)

[classylady]

Oh, why do I always feel the need to read and understand the math posts? I love math. To me, it's like putting pieces of a puzzle together and coming out with a beautiful picture (answer). Unfortunately, math is not my forte'. But, I love it all the same. Maybe in the next life my brain will be able to finally understand/comprehend it all.

[Dravin]

Okay, I see what you did and it makes sense. We've (the class and book) been differentiating with respect to dx but noting your process as far as I can see it just involves what variable we like and a slightly different path. Looking back I realize my fatal flaw. I was, for some reason, forgetting distance*weight and was just integrating weight.

Because taking 62.5*4pi(4-x)dx (kinda forgot to include the dx in my earlier post for volume, sloppy posting) as my weight and then using x as my distance you have the same equation with different variables. Ultimately ∫62.5*4pi(4-x)xdx or 250pi∫(4x-x^2)dx from 0 to 4. I suppose that is one of the fun things about math, multiple paths to the same answer (in some instances). Kinda reminds me of solving systems of equations, there were a couple tools in the kit, each would get you there in a lot of cases (though some more elegantly than others).

Hopefully I can keep things straight and remember how to do solids of revolution and cylindrical shells when the axis of rotation isn't x=0 or y=0 for my test in a few hours. For some reason I have a bear of a time 'compensating' for the 'shifted' axis of rotation. Shifting the graph (for instance X^2 rotated along y=-1 and capped by y=n is the same as x^2 + 1 rotated around y=0 and capped by some y=n+1 (Gotta adjust your interval of integration of course)) but that has downsides, mainly in the ease (or rather the lack there of) of setting up an equivalent for some functions. I think I went over an example last night that finally clicked with me but we'll see.

P.S. I want to kill this computer lab keyboard, h, t, e, y, and r, are really sticky. I feel like I'm trying to punish the keyboard to get it to perform correctly.

Edited December 2, 2011 by Dravin

[Vort]

Okay, I see what you did and it makes sense. We've (the class and book) been differentiating with respect to dx but noting your process as far as I can see it just involves what variable we like and a slightly different path. Looking back I realize my fatal flaw. I was, for some reason, forgetting distance*weight and was just integrating weight.

Because taking 62.5*4pi(4-x)dx (kinda forgot to include the dx in my earlier post for volume, sloppy posting) as my weight and then using x as my distance you have the same equation with different variables. Ultimately ∫62.5*4pi(4-x)xdx) or 250pi∫(4x-x^2)dx from 0 to 4. I suppose that is one of the fun things about math, multiple paths to the same answer (in some instances). Kinda reminds me of solving systems of equations, there were a couple tools in the kit, each would get you there in a lot of cases (though some more elegantly than others).

Hopefully I can keep things straight and remember how to do solids of revolution and cylindrical shells when the axis of rotation isn't x=0 or y=0 for my test in a few hours. For some reason I have a bear of a time 'compensating' for the 'shifted' axis of rotation. Shifting the graph (for instance X^2 rotated along y=-1 is the same as x^2 + 1 rotated around y=0 (Gotta adjust your interval of integration of course)) but that has downsides, mainly in the ease (or rather the lack there of) of setting up an equivalent for some functions. I think I went over an example last night that finally clicked with me but we'll see.

P.S. I want to kill this computer lab keyboard, h, t, e, y, and r, are really sticky. I feel like I'm trying to punish the keyboard to get it to perform correctly.

For some reason, when you wrote "cylinders" I thought you meant "cylindrical disks" rather than "cylindrical shells". If you use cylindrical shells -- which is what you are doing if you are integrating with respect to x -- it's much messier. For one thing, you need to figure out what to use for the distance through which you lift the cylindrical shell of water. It's not immediately obvious, since the distance to y=4 varies throughout the height of the cylindrical shell.

After a little thought, I'm sure you can just use the average distance of the shell's height to the top, or in other words, half the shell's height, which is (4-y)/2 = (4-¼x²)/2 = (2-x²/8).

Let's try it. The circumference of the shell at any position x is just 2πx, and the cylindrical height is 4-y = 4-¼x², so the total differential volume is:

dV = (2πx) (4-¼x²) dx

= π (8x-½x³) dx

and the weight is just ϱ dV, where ϱ is the density (which we consider to be uniform).

As we previously discussed, the (average) height you have to lift this shell is (2-x²/8), so the total differential work done in lifting that cylinder is

dW = (2-x²/8) ϱ dV

= (2-x²/8) (ϱπ) (8x-½x³) dx

= ϱπ [16x - x³ - x³ + (x^5)/16] dx

= ϱπ [16x - 2x³ + (x^5)/16] dx

Integrating from x=0 to x=4 yields:

W = ϱπ ∫ [16x - 2x³ + (x^5)/16] dx {from x=0 to x=4}

= ϱπ [8x² - ½(x^4) + (x^6)/96] {from x=0 to x=4}

= ϱπ [128 - 128 + 4096/96 - 0 + 0 - 0]

= 128 ϱπ/3

Substituting ϱ = 62.5 lb/ft³ gives

W = 8000π/3 ft-lbs

as before. But notice how much uglier it is to use cylinders here. Disks are just lots better, by far simpler and cleaner, none of this "average distance" garbage and other such uglification factors.

Good luck on the test. Hope everything goes well.

Edited December 2, 2011 by Vort
Clarification

[Dravin]

For some reason, when you wrote "cylinders" I thought you meant "cylindrical disks" rather than "cylindrical shells". If you use cylindrical shells, it's much messier.

Sorry, just my funny way of thinking. I think you parsed me correctly the first time; each disk is a cylinder of dx height.

[Vort]

Sorry, just my funny way of thinking. I think you parsed me correctly the first time; each disk is a cylinder of dx height.

That's fine, but if you're integrating using disks, I don't see how you can be using dx. The disks stack in the y direction, and their differential thickness is along y, not x. What am I missing here?

[Dravin]

That's fine, but if you're integrating using disks, I don't see how you can be using dx. The disks stack in the y direction, and their differential thickness is along y, not x. What am I missing here?

Quirky notation (or sloppy depending on however you want to view it) from examples from class and the text, the y-axis is basically being renamed. It's a hold over form the explanation of using Riemann sum set-ups to explain the process of calculating lifting a chain from the top of a cliff and from doing tanks where the only variable is y but is being called x (such as a cylindrical tank, horizontal distance is actually on the y axis but it's getting called x).

IIRC the instructor actually commented that some people are very careful to the coordinate things to the Cartesian planes and that he wasn't one of those people. I can see how that can have it's downside (that is not being particular) but it wasn't a handicap to solving this particular problem, the boneheaded mistake of forgetting weight*distance was.

[Vort]

Good enough. I guess that I'm one of those "careful" people, at least in this. In my experience, if you are not very careful when setting up your problem, including naming the axes, you are asking to get confused and screw things up. But hey, if things are working for you, then don't fix what ain't broke.

[Dravin]

Good enough. I guess that I'm one of those "careful" people, at least in this. In my experience, if you are not very careful when setting up your problem, including naming the axes, you are asking to get confused and screw things up. But hey, if things are working for you, then don't fix what ain't broke.

Previously it's all been simple with one axis change and everything easily relating to it (such as the 'boxes' of a triangular trough, or the height of a cylindrical tank) and x could very well have been C or Z. I can see the value in it though, a little caution goes a long way, being sloppy can cause issues but being careful rarely does.

BTW, I appreciate your help with my calculus questions. Thank you.

Edited December 2, 2011 by Dravin

[Vort]

Happy to help. Seriously. I haven't done calculus in years, and it's fun to dust off the ol' brain and give things a whirl. Plus, I like your posts elsewhere, so I'm happy to help out if I can. Hope things really have been of some help, and again, good luck on your tests.