Find the area by using disk method

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    Area Disk Method
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Discussion Overview

The discussion revolves around finding the area of a region R using the disk method, particularly focusing on the mathematical formulation and integration involved in the process. Participants are exploring the application of the disk method in relation to volumes generated by rotating a function around a line.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant seeks assistance in solving for the area R, indicating previous unsuccessful attempts.
  • Another participant describes the volumes V1 and V2 in relation to the x-axis and the line y=1, respectively.
  • A mathematical expression is provided, showing the relationship between the volumes and integrals involving the function f(x) over the interval [1, 4].
  • There is a query about the origin of the term (f(x)-1)^2 and the reasoning behind including -1 in the expression.
  • A repeated request for clarification on the derivation of (f(x)-1)^2 and its significance in the context of the disk method.
  • A question is posed regarding the radius of a disk when rotating the region R about the line y=1.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus, as participants express confusion and seek clarification on specific mathematical expressions and concepts related to the disk method.

Contextual Notes

Participants have not fully resolved the mathematical steps involved in the integration process, and there are unresolved questions regarding the formulation of the volume expressions.

jaychay
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The problem is to solve for the area R.

Can you please help me ?
I have tried to do it many times.

Thank you in advice.
 
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Volume V1 is on x - axis
Volume V2 is on y=1
 
$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$

$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$

$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$

note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?
 
skeeter said:
$\displaystyle V_1 - V_2 = \pi \int_1^4 [f(x)]^2 \, dx - \pi \int_1^4 [f(x)-1]^2 \, dx = 4\pi$

$\displaystyle \int_1^4 [f(x)]^2 \, dx - \int_1^4 [f(x)]^2 - 2f(x) + 1 \, dx = 4$

$\displaystyle \cancel{\int_1^4 [f(x)]^2 \, dx} - \cancel{\int_1^4 [f(x)]^2 \, dx} + \int_1^4 2f(x) - 1 \, dx = 4$

note $\displaystyle \int_1^4 f(x) \, dx = R + 3$
can you finish?
Untitled 2.png


Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?
 
jaychay said:
https://www.physicsforums.com/attachments/10787

Can you explain to me please where did (f(x)-1)^2 come from ? and why you have to put -1 behind f(x) ?

rotating R about the line y = 1 using disks ... what is the radius of a disk in this case?
 

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