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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2006-02-20 10:17:17

solids defined by cross sections. Those I've done. They're very cool.

2006-02-20 07:02:09

You've got me ryos.  I don't even know why you would bother using a third coordinate.  It looks like the figure you have there can be described by boundaries in just the x and y axis.  The way you are describing it, I don't even think that revolutions of solids is appropriate.  Three dimensional calculus is seperate from what we are talking about here.  The formula we are using here is derived based on a two dimensional analysis. I would dig into vector or multivariable calculus to solve it in the terms that you want to use.

2006-02-20 06:28:27

And if you have a shape of known cross section, the integral becomes much simpler.

Like this paraboloid, for example. The equation is z = -(x² + y²) + 10. The cross section is a circle of radius √[-z + 10] = x² + y². The circumference is then 2π√[-z + 10] = x² + y².

You then integrate the circumference multiplied by the differential of z, dz.

In coming up with this example, I've realized a deficiency in my instruction that I'm sure will be filled at a later date, but I want to know now. big_smile How would I set up said integral? Always before when I've done problems like this, they've given me a nice neat expression for r. Now...I don't know.

2006-02-20 05:59:53

Yeah usually its pretty worthless but sometimes it can be usefull in drawing simple shapes quickly.

2006-02-20 05:52:46

Really?  Okay, I can play around with that.  I haven't tried to much with paint even though it has been out forever.  Honestly, I really never used it for much more than image manipulation, because the tools are soooo limited.  I obviously am too cheap to go out and buy a decent software program.  Thanks anyway.

2006-02-20 05:45:07

lol. MS paint. I just drew two ellipses, connected the lines and filled in color to make it look how I wanted.

I'm still not quite sure how that works. But I'll turn it over in my head a while to see if I can get it to make sense.

2006-02-20 05:34:04

Yes you will and the formula I posted above multiplies the the average circumference of the rim by the change in arc length over that same distance.  Note that like the disk or washer method for calculating volumes that the formula above rotates about the same axis as the variable!

Oh, how do you generate those images that you use.  I am behind the times and would love to make such diagrams.  I understand the uploading and such, just not the creation of the image itself.  Is there a piece of software that you would recommend to me?

2006-02-20 05:24:08

this is what I'm seeing. If for instance you wanted to find the surface area of a sphere, sooner or later your gonna run into this piece:

2006-02-20 05:21:05

Think of the arc length as a "height" and the circumference as a "width".

2006-02-20 05:16:40

circumference times arclength. I figured it'd be something like  that but how can that work? In some cases the arc portion is horizontal and has thickness. :-/

2006-02-20 05:06:47

mikau, the integral for a surface of revolution is;

  ∫2πf(x)√(1 + [f'(x)]²) dx

  It is basically a circumference times the arc length.

  I will say that the original post seemed more like a related rate problem though.

2006-02-20 05:01:27

My book never discussed surface area of solids of revolution. I have a few idea's for how it might be accomplished but I've never tried. Maybe I should...

Anyways, yeah that pic didn't come through. Give it another shot if you will.

2006-02-20 03:46:40

Try uploading the image again Ansette.

2006-02-20 03:45:34

We have been given the problem of trying to find the minimum surface area given two variables in such a shape as the one in the image uploaded. any ideas?

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