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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    cross sectional area of a hollow cylinder of inner radius \[x\] of thickness \[dx\] will be \[2\pi xdx??\] My attempt: |dw:1442644796376:dw| If we open this we get a trapezium if I'm not wrong |dw:1442644914454:dw| Then we apply area of trapezium formula \[A=\frac{a+b}{2}.h\]\[\therefore dA=\frac{2\pi(x+dx)+2\pi x}2{}.dx\]\[\therefore dA=(\pi(x+dx)+\pi x)dx\]\[dA=(\pi x+\pi dx+\pi x)dx\]\[dA=(2\pi x+\pi dx)dx=2\pi x dx+\pi(dx)^2\] Since dx is extremely small, (dx)^2 will be even smaller and can be ignored \[dA=2\pi x dx\]

  2. dan815
    • one year ago
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    cross sectional area if u slice it how?

  3. IrishBoy123
    • one year ago
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    or \(dA = \pi (x + dx)^2 - \pi x^2 = 2 \pi x \, dx\)

  4. anonymous
    • one year ago
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    Oh I was intrigued with opening it, I forgot about the area formula, either way the term containing (dx)^2 is ignored right

  5. IrishBoy123
    • one year ago
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    and it's brill they agree!!

  6. anonymous
    • one year ago
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    lol, math always agrees!!

  7. anonymous
    • one year ago
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    cheers

  8. IrishBoy123
    • one year ago
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    the dx^2 disappearing is the ghost of departed quantities https://en.wikipedia.org/wiki/The_Analyst

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