Find the area by using disk method

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    Area Disk Method
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SUMMARY

The discussion focuses on calculating the area R using the disk method, specifically when rotating a region about the line y = 1. The volume equations provided, V1 and V2, utilize the integral expressions involving the function f(x) and its transformations. The key equation derived is V1 - V2 = 4π, leading to the conclusion that the integral of f(x) from 1 to 4 equals R + 3. The transformation (f(x) - 1)^2 is essential for determining the volume of the disks formed during rotation.

PREREQUISITES
  • Understanding of integral calculus, specifically definite integrals.
  • Familiarity with the disk method for calculating volumes of revolution.
  • Knowledge of function transformations, particularly (f(x) - c) for vertical shifts.
  • Basic skills in manipulating algebraic expressions involving integrals.
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  • Study the disk method in detail, focusing on applications in volume calculations.
  • Learn about function transformations and their implications in calculus.
  • Explore examples of calculating areas and volumes using integrals.
  • Investigate the relationship between integrals and geometric interpretations in calculus.
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Students and educators in mathematics, particularly those studying calculus and integral applications, as well as professionals involved in mathematical modeling and analysis.

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?
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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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