Volume of Solids of Revolution

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Homework Help Overview

The discussion revolves around finding the volume of a solid of revolution formed by rotating the area under the curve y = cos(x) from x = 0 to x = π around the line x = 1. Participants express confusion regarding the setup and the implications of the rotation.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the implications of rotating the area around the line x = 1, questioning whether the area can be simplified by focusing on only part of the curve. There is uncertainty about the correct interpretation of the problem and the symmetry of the region involved.

Discussion Status

The discussion is active, with participants raising questions about the problem's setup and the nature of the volume created by the rotation. Some suggest that the original poster may have misunderstood the problem, while others are considering the implications of ignoring certain parts of the curve.

Contextual Notes

There is a lack of clarity regarding the correct axis of rotation, with suggestions that the problem may involve rotating around y = 1 instead of x = 1. Participants are also discussing the unusual characteristics of the problem, which may affect their approaches.

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Find the volume of the solid st,
1. y=cos x , y= 0 in [0,pi] ; Rotated around x=1




2. I am slightly confused, I see that the area will double around twice so I can just use the left half of the curve. I am just not sure how to do so.
 
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Are you sure that you copied this problem correctly. The line x = 1 does not divide the region between y = cosx and y = 0 into two equal halves, so rotating it around the line x = 1 makes a complicated volume of revolution. Could it be that you're supposed to rotate the region around the line y = 1?
 
Yes, not 2 equal halves but the rotation around the x=1 axis would cover the right hand side of the curve in rotation and double over so can't we ignore that piece?
 
No, I don't think so. The part of the region under y = cosx on [0, 1] would cover the part on [1, pi/2], so I suppose you could ignore that part. The part of the curve on [pi/2, pi] is below the x-axis. The part above [0, 1] is above the x-axis. This is a very unusual problem.
 

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