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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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 \[dA=2\pi x dx\]
cross sectional area if u slice it how?
or \(dA = \pi (x + dx)^2 - \pi x^2 = 2 \pi x \, dx\)

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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
and it's brill they agree!!
lol, math always agrees!!
cheers
the dx^2 disappearing is the ghost of departed quantities https://en.wikipedia.org/wiki/The_Analyst

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