CENTROID OF PLANE FIGURES BY INTEGRATION Y π₯π ππ΄ π¦π 0 X Consider a plane figure as shown in fig, for which the centroid is required. First of all take an elementary strip and find its area. Let it be dA. Let π₯π and π¦π are the centroidal distance of the elementary strip from the reference axes. Then the first moment of the elemental area about x and y axes are π¦π ππ΄ and π₯π ππ΄ respectively. Moment of the whole area about 0π¦ axis = π₯π ππ΄ π΄π₯ = π₯π ππ΄ π₯π ππ΄ π΄ Moment of the whole area about 0π₯ axis= π¦π ππ΄ π₯= |||ππ¦ π΄π¦ = π¦π ππ΄ π¦π ππ΄ π΄ π¦= CENTROID OF A RECTANGLE TO FIND π Y π β 0 π₯ X ππ₯ Consider a vertical rectangle strib of thickness βππ₯β parallel to the height at a distance of π₯ from reference axis0π¦. Area of the strip ππ΄ = ππ₯. β β΄ Area of the rectangle, π΄ = ππ΄ = ππ₯. β π =β ππ₯ 0 = β π₯ π0 π΄ = βπ Using the result π₯ = here π₯ = π₯ + ππ₯ 2 π₯π π΄ ππ΄ ; but βππ₯β is very small, hence π₯π can be taken as equal to π₯. β΄π₯= π₯ππ΄ = π΄ π 0 π₯ βππ₯ β = βπ βπ 1 π₯2 = π 2 π π₯ππ₯ 0 π 0 1 π2 π π₯= = π 2 2 To find π Y Area of the strip, ππ΄ = πππ¦ Area of Rectangle, π΄ = β 0 ππ¦ πππ¦ π¦ π΄ = π π¦ β0 π΄ = πβ π¦π Using result π¦ = here π¦π = π¦ + ππ¦ 2 π΄ π ππ΄ β 0 ; dy is very small, hence π¦π = π¦ β΄π¦= β π¦πππ¦ 0 πβ 1 π¦2 = β 2 = 1 β2 β 2 π¦= β 2 β 0 X
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