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6.4 Second Moment of Area

The first moment of area (about the y axis) was

{M}_{x} ≈{\mathop{∑ }}_{a}^{b}xδA ={ \mathop{∑ }}_{a}^{b}xf(x)dx →{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}xf(x)\kern 1.66702pt dx.

Similarly second moment is same but with {x}^{2} instead of x,

\begin{eqnarray*} δx& =& {\mathop{∑ }}_{a}^{b}{x}^{2}\quad , %& \\ δA& =& {\mathop{∑ }}_{a}^{b}{x}^{2}f(x)dx →{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}{x}^{2}f(x)\kern 1.66702pt dx\quad .%& \\ \end{eqnarray*}

Example 6.9: 

Find the second moment of area under y = 1 + x + {x}^{2} about the y axis from x = 0 to x = 2.

Solution: 

\begin{eqnarray*} δx& =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2}{x}^{2}(1 + {x}^{2} + {x}^{3})\kern 1.66702pt dx%& \\ & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2}({x}^{2} + {x}^{3} + {x}^{4})\kern 1.66702pt dx %& \\ & =& \bigg {({{x}^{3}\over 3} + {{x}^{4}\over 4} + {{x}^{5}\over 5} \bigg )}_{0}^{2} %& \\ & =& {8\over 3} + {16\over 4} + {32\over 5} %& \\ & =& {40 + 60 + 96\over 15} %& \\ & =& {196\over 15} = 13 {1\over 15}. %& \\ \end{eqnarray*}

Note:  To find second moment about x axis is more complicated:

δy ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}{1\over 3}{f(x)}^{3}\kern 1.66702pt dx.

This will not be done here.

Note:  Recall that first moments are used in calculating centroids which are related to centres of mass.

Second moments are used in calculating moments of inertia of flat planes.

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