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piscean

  • one year ago

question: the volume of a disk in solution 8 is calculated as 2 {\pi} r dx. Should it include height as well if you use volume for solids of revolution when you revolves around y axis. question:http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4a.pdf solution: http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4asol.pdf

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  1. Waynex
    • one year ago
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    Certainly. Suppose the density is uniform in the height which we will let be the z axis. At every value for z, the density is represented by r^2. Then we can simply multiply the weighted area by the height. If somehow the density varies according to the z value, r^2 +sqrt(z) for example, then you would need to integrate over the z variable in a second round of integration. But that is a multivariable calculus problem. At that point though we probably want to use cylindrical or spherical coordinates instead of x, y, z. If that doesn't mean much to you, it will when you get to 18.02 ^__^

  2. Stacey
    • one year ago
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    It should include a height.

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