Solve Volume Problem: Find Region Bounded by Line & Curve

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SUMMARY

The discussion focuses on calculating the volume of the solid formed by the region bounded by the line y=x and the curve y=x², revolving around the line x=1. Participants suggest using the Washer Method and the Shell Method for integration. The correct integral for the Washer Method is identified as π[(1-y)² - (1-y^(1/2))²] dy from 0 to 1, leading to a final volume of π/6. The importance of correctly identifying the larger and smaller radii in the Washer Method is emphasized.

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  • Understanding of integral calculus
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  • Knowledge of the Shell Method for volume calculation
  • Ability to manipulate and simplify algebraic expressions
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  • Study the Washer Method in detail, focusing on radius identification
  • Learn the Shell Method and its application in volume problems
  • Practice solving volume integrals involving curves and lines
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Students studying calculus, particularly those focused on volume calculations, as well as educators looking for examples of applying the Washer and Shell Methods in real-world scenarios.

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Homework Statement



Find the volume of the soild formed.
The region bounded by the line y=x, and the curve y=x^2. about the line x=1

Homework Equations



Either Washer or Shell

The Attempt at a Solution



I want to use washer, but have to switch to dy.
I want to use shell, but don't think it will work.

I don't think this is right, but here is what I have.

integral of pie[(1-y^(1/2))-(1-y)]dy from 0 to 1.

Thanks, please help.
 
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Ok for a washer the area is pie(r1^2 - r2^2), where r1 is the larger radius and r2 is the smaller radius. So I think you just need to take a closer look at which is the larger and smaller radius and remember to square them too.

OTOH if you used shells you might not have to deal with a square root term in your integral.
 
the about x=1 is what mess me up. i meant to spuare all the R's but forgot, because this is hard to see.

integral of pie[(1-y)^2-(1-y^(1/2))^2]dy from 0 to 1.

this is right now?

I got pi/6, it looks right, but is it?
 
Last edited:
Looks correct to me.
 

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