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anonymous

  • 5 years ago

name the solid obtained by rotating the set of points (x,y) such that (x-4)^2 +y^2 < or =1 around the y-axis. (set up, but do not evaluate, an integral whose value is the volume of this solid. )

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  1. watchmath
    • 5 years ago
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    Look at here: http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd5dd7ad95c8b0bd2f35cc4

  2. anonymous
    • 5 years ago
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    its kinda confusing to me :(

  3. anonymous
    • 5 years ago
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    there is another way to do this I think. You look at the area of the circle, and multiply that by r*d(theta). That is your differential Volume element. Then you integrate from 0 to 2*Pi. It has the same result as watchmath's method. \[\int\limits_{0}^{2\pi}\pi*1^{2}*4d \theta\]

  4. anonymous
    • 5 years ago
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    where r is the radius to the center of the circle from the origin, and youre integraing about the y axis

  5. watchmath
    • 5 years ago
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    So we can think of this \(rd\theta\) as the thickness of this differential volume?

  6. anonymous
    • 5 years ago
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    i think so. it is the arclength of one small piece of the volume which would be the thickness of the differential element. Then adding these all up around the y axis gives the volume.

  7. watchmath
    • 5 years ago
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    Jojo, are you familiar with the cyllindrical shell method?

  8. anonymous
    • 5 years ago
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    watchmath is that method sort of a way to slice the circle into vertical strips and add up each shell over the diameter of the circle? It's been a while since I did these problems.

  9. watchmath
    • 5 years ago
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    Yes, that's correct.

  10. anonymous
    • 5 years ago
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    thank you guys! but please help me with other my review questions?? :(

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