Volume of Solid of Revolution with Disk Method: y=5x^2, x=1, y=0 about x-axis

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Discussion Overview

The discussion revolves around calculating the volume of a solid of revolution formed by rotating the area bounded by the curve y=5x², the line x=1, and the line y=0 about the x-axis. Participants explore the application of the disk method for this problem, including the setup of the integral.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to start the problem and finding the integral.
  • Another participant describes the disk method, suggesting that the volume can be calculated by integrating πy² dx from 0 to 1.
  • A participant attempts to apply the disk method but arrives at an answer that they believe is incorrect, questioning where they went wrong in their calculations.
  • Another participant points out that the expression y² is not equal to 5x³/3 and emphasizes the need to integrate πy² dx, clarifying that y=5x² should be used in the integral.
  • One participant expresses confidence that the disk method in terms of x should be effective for solving the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach, as there are differing interpretations of the integral setup and calculations involved. Some participants challenge the correctness of earlier claims without resolving the disagreement.

Contextual Notes

There are unresolved aspects regarding the correct formulation of the integral, particularly in relation to the expression for y² and the limits of integration. The discussion reflects varying levels of understanding of the disk method and its application to this specific problem.

Bricenob
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Member warned about posting homework questions in a non-homework section
I was wondering if anyone could help me with this. I'm stuck and not sure where to start/how to go about it and finding the integral as well...

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=5x^2,x=1,y=0, about the x-axis
 
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Imagine a line drawn perpendicular to the x-axis up to the give parabola. Rotated around the x-axis, it will sweep out a disk with area \pi y^2 so, imagining a thickness of "dx", volume \pi y^2 dx. Integrate that from 0 to 1.
 
HallsofIvy said:
Imagine a line drawn perpendicular to the x-axis up to the give parabola. Rotated around the x-axis, it will sweep out a disk with area \pi y^2 so, imagining a thickness of "dx", volume \pi y^2 dx. Integrate that from 0 to 1.
so when I do that.. I come up with (pi)(int 0-->1)(5*((x^3)/3)= (pi) (int. 0-->1) 5(1^3)/3= 5pi/3. But it says this answer is incorrect. Where am I going wrong?
 
If y= 5x^2 then y^2 is NOT equal to "5x^3/3"!

Or did you mean that 5x^3/3 is the result of integrating x^2 dx? (If so do NOT continue to write "int. 0-->1"!)

That would be the integral of \pi x^2dx but that is NOT what you should be integrating! You should be integrating \pi y^2 dx. And, of course, y= 5x^2.
 
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I think the disk method in terms of x should work.
 

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