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Volumes by the Method of Cross Sections



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Tue Nov 28, 2006 4:53 pm
Dream Deep says...



*wherever the variables a or b are mentioned, they refer to the integral existance theorem, where 'If f is continuous on [a,b], then f is integrable on [a,b]'.

*dark blue text represents the equation or theorem, dark red represents notes in the midst.


The directions are:

Find the volume of the solid that is generated by rotating around the indicated axis the plane region bounded by the given curves.

Indeed. The problem:

y=x^2, y=0, x=1; the x-axis


So that's all I get to work with. And I'm not doing such a fine job of finding the volume of the solid. If it helps at all, the central theorem of the lesson was "The Definition of Volume by Cross Sections":

If the solid R lies alongside the interval [a,b] on the x-axis and has continuous cross sectional area function A(x), then its volume V=v(R) is V=∫ (where 'b' and 'a' remain in variable form) A(x) dx.

Any thoughts? o0
  





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Wed Nov 29, 2006 3:36 pm
Incandescence says...



DD--


How is it you are setting up your problem?

Take the curve y=x^2, y=0, and y=1. You should see that the volume you are solving for is a (funky-shaped) "ring." So the question is now: which axis are you rotating about? If you rotate about x, you use the dx integral; y, the dy integral.

So our solution is of the form: V(x)=∫A(x)dx.

How do we define our area function?

A(x)=pi*r^2. Well, let's look to our problem statement. We know the radius here is not a constant, but a variable that varies according to the function y=x^2. So let's substitute by A(x)=pi*(x^2)^2=pi*x^4. Now let's re-set up our integral.

V(x)=pi*∫_0^1(x^4)dx.


Don't you know how to solve that?
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Wed Nov 29, 2006 5:01 pm
Dream Deep says...



Thanks a lot for the help, Brad... the answer key informs me that the outcome should be pi/5... o0

I think I'll work off of your explanation to simplify it out. My problem was set up in a real mess - you don't want to know, it would put you in cardiac arrest.

Thanks again, Brad, for the pointers. ^_~
  





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Wed Nov 29, 2006 8:09 pm
Griffinkeeper says...



Then it is around the x axis.

∫A(x)dx.

∫ pi (x)^2 dx

We have the lines y=0 and y=x^2, which means that we have a parabola and with the lower boundary of the x-axis.

We're given that x=1. This means that x goes from the y-axis to x=1.

Since it is in terms of x, then the integral goes from 0 to 1.

Therefore.

∫ pi (x^2)^2 dx. (0<x<1)

pi ∫ x^4 dx (0<x<1)

(pi x^5)/5 (0<x<1)

((pi*1^5)/5- (pi*0^5)/5)

Simplified, we get pi/5.
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Wed Nov 29, 2006 10:32 pm
Dream Deep says...



Hey, thanks a lot, Grif. ^_~

Without the two of you I would be very, very dead concerning this lesson. Calculus and I don't like each other. Thanks for the aid. ^_^
  





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Sat Dec 16, 2006 7:57 am
Mr. Everyone says...



You could most probably find help on wikipedia, or do forum search, =) good luck and =)happy reading(=
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