How Do You Apply the Washer Method for Calculating Volumes of Revolution?

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SUMMARY

The discussion focuses on applying the washer method to calculate volumes of revolution for the region bounded by the curve y = √x, the line y = 2, and the vertical line x = 0, over various axes. The volume calculations provided include revolutions around the x-axis, y-axis, the line y = 2, and the line x = 4, yielding results of 42.265, 8.3775, 8.3775, and 46.9144 cubic units respectively. Key feedback highlighted the importance of using parentheses in the integral expressions to avoid confusion in calculations.

PREREQUISITES
  • Understanding of the washer method for volume calculation
  • Familiarity with integral calculus
  • Knowledge of functions and their graphs, specifically y = √x
  • Ability to manipulate and evaluate definite integrals
NEXT STEPS
  • Study the washer method in detail, focusing on its application in different scenarios
  • Practice evaluating definite integrals involving square roots and polynomials
  • Learn about the disk method as an alternative to the washer method
  • Explore common pitfalls in volume calculations, particularly with parentheses in integrals
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Students studying calculus, educators teaching volume calculations, and anyone seeking to master the washer method for finding volumes of revolution.

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Homework Statement


Use the washer method to find the volume of hte solid generated by revolving the regions bounded by y= root(x) and the lines y =2 and x=0 about the x-axis , y-axis , y= 2 and x=4


Homework Equations



V= Pi integral (0,4) (R(x)^2 - r(x)^2) dx

The Attempt at a Solution



I tried to use the equation above but I totally got confused about r(x)^2 as from what I know that the washer method means you have a small radius that you will subtract from the large curve which is in this case root(x) but there is none!

Thank you very much for your time, I would appropriate if you could explain how you solve this question so I could learn
 
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I tried to solve these problems in a different way but I'm not sure if I'm on the right track.

If it revolved over the x-axis then the volume will be
integral (0,4) Pi (2)^2 - (root (x))^2 dx = 42.265 units^3

If it revolved over the y-axis then the volume will be
integral (0,2) Pi (y)^2 - 0 dy = 8.3775 units^3

If it revolved over the line y=2 then the volume will be
integral (0,4) Pi (-root(x)+2)^2 dx = 8.3775 units^3*****This is the one that I'm not really sure if its the right answer or Not so please HELP!

If it revolved over the line x=4 then the volume will be
integral (0,2) Pi ((4)^2 -(-y^2+4)^2) dy = 46.9144 units^3
 
karamsoft said:
I tried to solve these problems in a different way but I'm not sure if I'm on the right track.

If it revolved over the x-axis then the volume will be
integral (0,4) Pi (2)^2 - (root (x))^2 dx = 42.265 units^3

\int_0^4 \pi\left(2^2-(\sqrt{x})^2\right)\,dx

You're missing a set of parentheses.

karamsoft said:
If it revolved over the y-axis then the volume will be
integral (0,2) Pi (y)^2 - 0 dy = 8.3775 units^3
You missed the parentheses here too, but the numerical answer is OK, because zero times pi is zero.

karamsoft said:
If it revolved over the line y=2 then the volume will be
integral (0,4) Pi (-root(x)+2)^2 dx = 8.3775 units^3
This looks OK.

karamsoft said:
*****This is the one that I'm not really sure if its the right answer or Not so please HELP!

If it revolved over the line x=4 then the volume will be
integral (0,2) Pi ((4)^2 -(-y^2+4)^2) dy = 46.9144 units^3
This looks OK too.
 

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