Calculating Volume of Revolution: Solving for Unknowns Using Equations

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Homework Help Overview

The discussion revolves around calculating the volume of revolution using integrals, specifically focusing on the evaluation of a definite integral involving a function defined in terms of \(x\) and \(y\). Participants are attempting to reconcile their results with a given answer, questioning the correctness of their methods and calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration process and the importance of including factors such as \(\pi\) in their calculations. There are attempts to clarify the setup of the integral and the evaluation of the volume element.

Discussion Status

The conversation is ongoing, with participants sharing their integrals and results. Some express confusion over discrepancies in answers, while others suggest potential errors in calculations. There is no explicit consensus on the correct approach or outcome at this stage.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a focus on ensuring that all factors in the volume calculation are accounted for, particularly the inclusion of \(\pi\).

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Homework Statement



i have done the part a, for b , i use the key in the (circled part equation ) into calculator .. my ans is also different form the ans given. is my concept correct by the way?

Homework Equations





The Attempt at a Solution

 

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One obvious point is that you are missing a factor of "[itex]\pi[/itex]". The area of a circle is [itex]\pi r^2= \pi y^2[/itex].
 
after adding pi, my ans is 2.80... the ans is 5.047746784, which part is wrong?
 
It would be a good idea here to use vertical line segments, otherwise you're going to have to set up multiple integrals. So leave everything as ##y(x)##, then:

##r_{in} = 0##
##r_{out} = 1 + \frac{1}{4x^2 + 1}##
##height = dx##

##dV = 2\pi(\frac{r_{in} + r_{out}}{2})(r_{out} - r_{in})(height)##

Integrating the volume element should give you the answer you want.
 
delsoo said:
after adding pi, my ans is 2.80... the ans is 5.047746784, which part is wrong?

delsoo, your method is fine and I seem to get the same definite integral as you (which gives the correct answer too). The definite integral you have to evaluate is:

$$\pi\int_0^{1/2} \left(1+\frac{1}{4x^2+1}\right)^2\,dx$$

If you drop the factor of ##\pi##, you should get 1.60675.

Use the substitution ##2x=\tan\theta## to make things easier.
 
Pranav-Arora said:
delsoo, your method is fine and I seem to get the same definite integral as you (which gives the correct answer too). The definite integral you have to evaluate is:

$$\pi\int_0^{1/2} \left(1+\frac{1}{4x^2+1}\right)^2\,dx$$

If you drop the factor of ##\pi##, you should get 1.60675.

Erm this is misleading.

The answer is indeed 5.04775 complements of wolfram:

http://www.wolframalpha.com/input/?i=integrate+2pi%28+%281%2B+1%2F%284x^2%2B1%29%29%2F2+%29%281%2B+1%2F%284x^2%2B1%29%29+from+0+to+1%2F2
 
Zondrina said:
Erm this is misleading.

Can you please explain to me how my statements are misleading? :)
 
Pranav-Arora said:
Can you please explain to me how my statements are misleading? :)

Your integrand is fine, it's just the answer you got I was worried about.
 
Zondrina said:
Your integrand is fine, it's just the answer you got I was worried about.

I must be missing something but what is the problem with the answer I wrote? Are you talking about "1.60675"? :confused:
 
  • #10
Pranav-Arora said:
I must be missing something but what is the problem with the answer I wrote? Are you talking about "1.60675"? :confused:

Yeah I wasn't sure why you wrote that.
 
  • #11
Zondrina said:
Yeah I wasn't sure why you wrote that.

Ah, I think I worded it poorly. What I meant was this:
$$\int_0^1 \left(1+\frac{1}{4x^2+1}\right)^2\,dx=1.60675$$
And I feel delsoo did some mistake while evaluating the above definite integral.
 

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