[Blindmath] volume of rotational solids (calculus)

Tue Mar 8 23:40:51 UTC 2011

Dear Alex,

   I think as blind individuals we would like most things to be purely
algebra, but that really isn't a safe or reasonable way to go with a
lot of math.  In this case I think that picturing the graph, though
difficult at first, will end up being the most useful tool for you in
the future.  If you plan to go any farther in math, such as to calc
III, it is necessary that you visualize bizarre graphs with ridiculous
configurations.  Your problems now, if solved with graphical
visualization, will set you up to be able to work with surface
integrals and a lot of the 3-dimentional figures you'll need in the
future.  In short, sorry, but what you least want to do is probably
the most helpful for the time being and for the future.

Sincerely,
   Ryan

On 3/8/11, Alex Hall <mehgcap at gmail.com> wrote:
> I can only say what my professor has explained to me. He said that the
> function starts at the origin and so will, when rotated about the y
> axis, form a bowl shape, as you said. He recommends using the Shell
> Method for this, which looks like:
> 2*pi*(integral)(p(x)*h(x))
> where h(x) is the function and p(x) is x, since it is the distance
> from the axis of rotation out to the "shell" (think of it as a
> cylinder) you are calculating for.
>
> However, he also said that the Washer Method can be used here. In this
> method, find the area with:
> pi*(integral)(R(x)-r(x))
> Note that the first r is capitalized and the second is not. The point
> here is that you find the area of the whole solid, bowl included, as
> though the bowl were filled in; this is the first r. You then subtract
> the volume of the bowl - the little r - to get the volume of the
> function. The only way he has given me to know what this second r
> should be equal to is to think about the function's shape and imagine
> it being rotated. It then becomes clear that the first r is equal to
> your larger point of integration, 2, while your second r is equal to
> the function itself. You then have:
> pi*(int)(2)-e^x), or pi*2x-e^x. This will give you the same area as
> the Shell Method would have, in theory, but the only way to know how
> to set it up is to picture the function to find just what both r
> values should be. Things get more complicated when you throw constants
> into the mix and start moving the function away from the origin, which
> is why I am hoping for a general format to follow to find what I am
> looking for without having to try to figure out the graph itself and
> what it looks like. Sorry if this is not too clear; we only started
> this last Wednesday, we meet three times a week, and Monday's class
> was cancelled, so I am still pretty new to it. I just know that trying
> to picture each graph is not always going to work.
>
> On 3/8/11, Salisbury, Justin Mark <SALISBURYJ08 at students.ecu.edu> wrote:
>> Also, where on the interval (0,2) is e^x undefined?
>>
>> Justin M. Salisbury
>> Undergraduate Student
>> The University Honors Program
>> East Carolina University
>> salisburyj08 at students.ecu.edu
>>
>> "It is the mark of an educated mind to be able to entertain a thought
>> without accepting it."    -Aristotle
>> ________________________________________
>> From: blindmath-bounces at nfbnet.org [blindmath-bounces at nfbnet.org] on
>> behalf
>> of Alex Hall [mehgcap at gmail.com]
>> Sent: Tuesday, March 08, 2011 5:32 PM
>> To: Blind Math list for those interested in mathematics
>> Subject: Re: [Blindmath] volume of rotational solids (calculus)
>>
>> Yes, calc 2, and yes, only continuous functions. The problem is that,
>> sometimes, there is a region that is undefined for the function. If
>> you have some exponential, call it e^x, from x=0 to x=2 rotated about
>> the y-axis, there is a region shaped like a bowl that you must account
>> for by subtracting it out. For a more complex equation, imagining the
>> shapes is not so easy, so I could not just know that there is a region
>> whose area must be subtracted out. I am looking for a non-graphing way
>> to do this. After all, as you said, sketches are just a visualization
>> tool; functions should not need to be sketched to be worked with.
>>
>> On 3/8/11, Salisbury, Justin Mark <SALISBURYJ08 at students.ecu.edu> wrote:
>>> Hi Alex,
>>>
>>>     Are you taking Calculus II?  If so, I'm pretty sure that they aren't
>>> expecting you to integrate any functions that are undefined.  I'm pretty
>>> sure you'll be integrating continuous functions.  If you're asking how to
>>> know which bound lies inside the other one, that's generally a matter of
>>> a
>>> distance from the axis of rotation.  I hope that you can mentally
>>> conceptualize which functions lie inside other ones.  Otherwise, you'll
>>> have
>>> to use a distance formula to the axis of rotation for points on each
>>> equation.  If you just have the understanding that you're calculating the
>>> distance between two or more equations and then rotating it to turn area
>>> into volume, you should have a thorough enough understanding of what
>>> you're
>>> doing.  If you want to contact me off-list for humorous real-life
>>> examples,
>>> feel free.  Sighted mathematicians really only use sketches to help
>>> themselves understand what they're doing.  Many advanced mathematicians
>>> never even draw the functions before beginning their computations.  I
>>> think
>>> you'll be just fine as long as you know that you're accomplishing and
>>> know
>>> which functions lie inside the others.
>>>
>>> Good luck!
>>>
>>> Justin
>>>
>>> Justin M. Salisbury
>>> Undergraduate Student
>>> The University Honors Program
>>> East Carolina University
>>> salisburyj08 at students.ecu.edu
>>>
>>> "It is the mark of an educated mind to be able to entertain a thought
>>> without accepting it."    -Aristotle
>>> ________________________________________
>>> From: blindmath-bounces at nfbnet.org [blindmath-bounces at nfbnet.org] on
>>> behalf
>>> of Alex Hall [mehgcap at gmail.com]
>>> Sent: Tuesday, March 08, 2011 4:30 PM
>>> To: Blind Math list for those interested in mathematics
>>> Subject: [Blindmath] volume of rotational solids (calculus)
>>>
>>> Hi all,
>>> This is not a request for help in finding this sort of thing. Rather,
>>> I am wondering if it can be done purely algebraically so I do not have
>>> to try to imagine the graph. Example:
>>>
>>> Find the volume of the solid formed by rotating the function y=x^2
>>> around the x-axis from x=0 to x=4.
>>>
>>> This one is a pretty simple example, and should be pi*x^5/5, I think.
>>> This is using the Disk Method, but what happens with the Washer method
>>> or the Shell method, where you might have space in the solid where the
>>> function is not defined? Currently, I have to try to imagine the graph
>>> to "see" the radius to use, any undefined portions, and so on. What I
>>> am wondering is if anyone has dealt with this and has found any way to
>>> do it all with algebra or some other non-graphical method. If so,
>>> please share! Thanks.
>>>
>>> --
>>> Have a great day,
>>> Alex (msg sent from GMail website)
>>> mehgcap at gmail.com; http://www.facebook.com/mehgcap
>>>
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>>
>>
>> --
>> Have a great day,
>> Alex (msg sent from GMail website)
>> mehgcap at gmail.com; http://www.facebook.com/mehgcap
>>
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>
>
> --
> Have a great day,
> Alex (msg sent from GMail website)
> mehgcap at gmail.com; http://www.facebook.com/mehgcap
>
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