[Blindmath] volume of rotational solids (calculus)

[Salisbury, Justin Mark]

I just ran the function that you described, and I am truly confused by your question.  Since you are in calc 2, you must be doing an actual integral.  

The region above the function is not included in the integral because an integral is the area beneath the curve.  This is a two-dimensional slice that will, when rotated, form a bowl-type shape.  The bowl-type shape is what you're trying to achieve; it is the solid of revolution.  You don't have to subtract out the bowl because it is what you're trying to find. 

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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