Solve Volume Problem: Find Region Bounded by Line & Curve

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Homework Help Overview

The problem involves finding the volume of a solid formed by the region bounded by the line y=x and the curve y=x^2, revolving around the line x=1. The subject area pertains to calculus, specifically volume calculations using methods such as washers or shells.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the washer method and express uncertainty about switching to dy. There is also consideration of the shell method, with concerns about its applicability. Questions arise regarding the identification of the larger and smaller radii for the washer method and the correct formulation of the integral.

Discussion Status

Some participants have provided guidance on the washer method, emphasizing the importance of correctly identifying the radii and squaring them. Others have expressed confusion about the setup due to the axis of rotation and have attempted to clarify their integrals. There is a mix of interpretations regarding the correctness of the integrals presented.

Contextual Notes

Participants note the challenge posed by the axis of rotation at x=1 and the need to ensure all terms are squared in the integral. There is an acknowledgment of the complexity involved in visualizing the problem.

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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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