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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 \[dA=2\pi x dx\]
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