Solve Volume of Solid: y=x, y=x^1/2, y=1

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Discussion Overview

The discussion revolves around calculating the volume of a solid formed by rotating the area bounded by the curves y=x, y=x^(1/2), and the line y=1 about the line y=1. Participants are attempting to solve this problem and clarify their approaches, focusing on the correct setup of the integral for volume calculation.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant states they attempted to solve the problem but arrived at a different answer than the book, which claims the volume is pi/6.
  • Another participant suggests pinpointing the specific problem in the approach taken.
  • A participant shares their work, using a formula involving the integral of the outer and inner radii, but arrives at a volume of 7/6 pi, expressing confusion about their mistake.
  • One reply advises integrating with respect to x and provides a corrected integral setup, indicating the inner and outer radii based on the curves.
  • Another participant challenges the correctness of the integrand provided, asserting that the rotation should be around y=1 and gives a revised integral expression for the volume.
  • A later response points out a sign error and clarifies that the participant had mistakenly rotated around x=1 instead of y=1, which contributes to the discrepancy in their results.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views on the correct setup for the integral and the interpretation of the problem. Disagreements exist regarding the correct radii and the axis of rotation.

Contextual Notes

Limitations include potential misunderstandings of the problem setup, the need for clarity on the axis of rotation, and the correct identification of inner and outer radii. Some mathematical steps remain unresolved.

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volume of solid:(

hi..I tried to solve it but ı couldn't .book says the answer is pi /6...please help me.

question is; y=x and y=x^1/2 about y =1
 
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Pinpoint your problem.
 
ok I showed my work..


I used this formula
integral Pı(outer r)^2-Pı(inner)^2



inside integral from y=0 to y=1 pi(1-y)^2 - pi(1-y^2)^2

my result is 7/6 pi..:( where am I wrong??
 
integrate it with respect to x. The inner radius is 1-\sqrt{x} and the outer radius is 1- x.

So you have \int_{0}^{1} \pi((1-x)^{2}-(1-\sqrt{x})^{2}) \; dx
 
Last edited:
Well, your integrand is utterly wrong!
You are to rotate the VERTICAL strip \sqrt{x}\leq{y}\leq{x} around y=1.
That gives you the integral:
V=\int_{0}^{1}\pi((1-x)^{2}-(1-\sqrt{x})^{2})dx=\pi\int_{0}^{1}(x^{2}-3x+2\sqrt{x})dx
 
Apart from being wrongly signed, you have rotated your figure around the line x=1, rather than around y=1.
That accounts for your discrepancy.
(Note, for example, that your INNER radius is the actual OUTER radius)
 

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