When n tends to infinity О”W becomes infinitesimally small and can be replaced as dW.

Let О”W 1, О”W 2, О”W 3 ,вЂ¦вЂ¦., О”W n be the elementary forces acting on the elementary elements. LetвЂ™s say the coordinates to be (x 1 ,y 1 ), (x 2 ,y 2 ), (x 3 ,y 3 )вЂ¦вЂ¦вЂ¦. We can also divide the entire region into letвЂ™s say n small elements. As shown in first figure we can easily represent the net force passing through the single point G. Further letвЂ™s assume the body has practical dimensions, then we can easily conclude that all elementary forces will be unidirectional and parallel.Д¬onsider G to be the centroid of the irregular lamina. Obviously every single element will experience a gravitational force towards the centre of earth. LetвЂ™s assume that it has been exposed to gravitational field. Further we will also study the surface area or volume of revolution of a line or area respectively.Д¬onsider the following lamina. These parameters simplify the analysis of structures such as beams. Various such parameters include centre of gravity, moment of inertia, centroid, first and second moment of inertias of a line or a rigid body. We can extend this concept in many ways and get the various equivalent parameters of a body, which could help us in dealing the situation directly on a rigid body rather than considering each individual particle of the rigid body. This point is known as centre of gravity. We equivalently represent the system of forces by single force acting at a specific point. It could be the same Cartesian x,y axes, we have selected for the position of centroid.In general when a rigid body lies in a field of force acts on each particle of the body. We must decide on the working coordinate system. In the next steps we'll need to find only coordinate y c. In order to take advantage of the shape symmetries though, it seems appropriate to place the origin of axes x, y at the circle center, and orient the x axis along the diametric base of the semicircle.Д«ecause the shape is symmetrical around axis y, it is evident that centroid should lie on this axis too. The coordinate system, to locate the centroid with, can be anything we want. X_c = rac Example 2: centroid of semicircle using integration formulasДerive the formulas for the location of semicircle centroid Composite AreasДЇor composite areas, that can be decomposed to a finite number n of simpler subareas, and provided that the centroids of these subareas are available or easy to find, then the centroid coordinates of the entire area x_c, y_c can be calculated through the following formulas: The application of the procedure will become clear with the examples later in the article. Integrate, substituting, where needed, the x and y variables with their definitions in the working coordinate system.Describe the borders of the shape and the x, y variables according to the working coordinate system.Called hereafter working coordinate system. It can be the same (x,y) or a different one. Select an appropriate, and convenient for the integration, coordinate system.Select a coordinate system, (x,y), to measure the centroid location with.The steps for the calculation of the centroid coordinates, x c and y c, through integration, are summarized to the following: Steps for finding centroid using integration formulas You may use either one, though in some engineering disciplines 'static moment' is prevalent. ' Static moment' and ' first moment of area' are equivalent terms.

Moment of inertia of a circle centroids