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Y'all might want to take a try at it. I'll post the proof in a couple of days

thanks vector22

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- Thread starter vector22
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- #1

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Y'all might want to take a try at it. I'll post the proof in a couple of days

thanks vector22

- #2

gb7nash

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err nevermind, misread the OP

- #3

jhae2.718

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It's a trivial double integral in polar coordinates...

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HallsofIvy

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It is, as jhae2.718 said, a trivial double integral in polar coordinates. It's a bit more fun to do it in xy- coordinates.

You might also want to prove that the volume of a sphere is [itex](4/3)\pi r^3[/itex] using

1) spherical coordinates. (trivial)

2) cylindrical coordinates. (a little harder)

3) Cartesian coordinates. (fairly hard)

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you can do it without polar coordinates.

y'all come back now

y'all come back now

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you can do it without polar coordinates.

y'all come back now

Ahhyup! Y'all can also do it with geometry as the limit of the areas of inscribed regular polygons, y'hear?

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HallsofIvy

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mathwonk

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It also fascinated me to try to generalize and compute the volumes of higher dimensional spheres. when you try it in a way directly analogous to the computation of the usual three dimensional sphere (by slicing) you get an integral with a square root or a n/2 power in it and those are harder than the easy one for the sphere. I.e. the even dimensional spheres are harder by this method than the odd dimensional ones.

Notice in particular that doing a disc (dimension 2) you get a harder integral than for a sphere, but we think we know how to do it because we think we know about trig substitutions and arcsins and so on. Reflect however that even the definition of the sin and arcsin functions actually depends on knowing properties of circles like arc length, which are more difficult than what we are doing. So one should not allow the area of a circle to be computed by integrating (1-x^2)^(1/2) which uses more sophisticated concepts than what they are being applied to.

I ultimately noticed that the easy way to do volumes of 4 spheres and 5 spheres and so on is to use cylindrical shells, since then one gets a simpler integral. This arose when I noticed that the work calculation for pumping water out of a hemispherical pool is actually equivalent to calculating the volume of a 4 sphere by cylindrical shells. (For a disc, this corresponds to sweeping out the disc by a family of expanding circles, or polar coordinates.)

Of course if you want to cheat and throw all the difficulty into the coordinates for the integral, it makes more sense to do the volume of a ball by spherical integration, not polar. I.e. sweeping it out by a family of expanding spheres. But how do you generalize these to dimension 4?

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Being a 2nd/3rd year community college science student (transferring this fall with every intention to major in pure Math) I wanted to ask when you say "to prove" something, do you mean an actual formal proof composition? Or is just some symbolic demonstration (ie like much of my homework, beginning with some kind of equation and then applying whatever math techniques to gradually transform how the equation is written to show that the original statement is true ) Or am I really displaying my ignorance and are these the same thing?

I ask because my Calculus text (Stewart) has a few problems in it asking to "prove" something, and I have had absolutely no formal proof education to this point and am terrified to be exposed to it for the first time this fall. However the solution for these problems appear to be nothing more than just that, a solution (no axioms or theorems are stated - nor any confusing symbols used which must certainly be mandatory! ).

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gb7nash

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Some visual aids will be helpful for those who want a detailed explanation of this. Get 2 metal washers with the same outer diameter but with different size holes. The dimensions of a flat metal washer are:

The thickness

The outer circumference

The inner circumference

The width of the metal portion (make that equal to ∆r)

For this explanation we are not concerned with the thickness of the washer.

You have two washers with different size holes (outer diameters the same) You will find out that the approximate area of the washer with the larger hole is closer to the actual area as compared to the washer with the smaller hole. The approximate area of the washer will approach the actual area of the washer as ∆r gets smaller.

The outer circumference = 2πr

The approx area of a washer is: (the area of the metal surface)

∆A = 2πr ∆r

Change ∆r to dr and integrate to find the exact area of a circle.

2π is outside the integration symbol as they are constant.

A = 2π∫r dr = 2πr^2 / 2 = πr^2

Can’t get any simpler than that.

- #12

disregardthat

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I confess to being one of the few who finds those problems interesting. I like the area of a circle proof by limits of triangles, since the calculus proofs assume things about trig functions and about the meaning of pi that are swept under the rug, and at least as difficult as what is being proved.

Yes, I really agree with you on this one. Often we even define the trigonometric functions as e.g. solutions of differential equations, or as power series - entirely disconnected to the geometric feature of them. This leaves the work to connect the missing dots.

In my opinion, assuming sin(x) is defined geometrically, the fundamental thing about the trigonometric functions is that [tex]\lim_{x \to 0} sin(x)/x = 1[/tex] when x is measured by sin in radians. This is the core of the proof of the area formula for a circle done by inscribing and circumscribing polygons. If sin(x) is defined geometrically, this is also a necessary step in proving the formula by integral calculus. However, the proofs I've seen of [tex]\lim_{x \to 0} sin(x)/x = 1[/tex] is actually using the formula for the area of a circle...

The point is:

Defining the trigonometric functions as power series or differential equations is not enough, you will have to prove the necessary links between the geometry and the definitions. Secondly, if you define sin(x) the classical way as opposite side divided by the hypotenuse of a right angled triangle, you will have to prove that [tex]\lim_{x \to 0} sin(x)/x = 1[/tex] without use of the area formula for a circle to do the necessary calculus to prove the formula by integration or inscribing/circumscribing polygons.

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Also, I too find these proofs of "simple" things very interesting.

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disregardthat

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Also, I too find these proofs of "simple" things very interesting.

Well, proving A= pi*(radius)^2 and A=1/2*(circumference)(radius) is pretty much the same thing, you haven't assumed anything more or less in either case. Don't be surprised if you will have to use the constant (circumference)/(2(radius)) a lot in the latter case...

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I only showed a proof for the area of a circle. I proved the area of the circle could be done by using the washer method.

- #16

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Well, proving A= pi*(radius)^2 and A=1/2*(circumference)(radius) is pretty much the same thing, you haven't assumed anything more or less in either case. Don't be surprised if you will have to use the constant circumference/(2(radius)) a lot...

In the first case you've assumed the defined constant pi, in the second case you haven't assumed the defined constant pi.

- #17

disregardthat

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In the first case you've assumed the defined constant pi, in the second case you haven't assumed the defined constant pi.

Why do you think I haven't "assumed the definition of pi" in the second case? That's not the kind of assumptions I'm talking about anyway.

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