in this video we will focus on how to determine the centroid of an area which is the geometric center of an area from the previous video we learned how to theoretically determine the location of the center of gravity for a rigid body with respect to a given coordinate system xbar Y Bar and z bar and we also know that weight equals to mass * G which is the gravity gra itational acceleration and for the General near Earth situation it can be considered as a constant about 9.81 m per second squared therefore the constant G can be canceled out from the numerator and the denominator of this equation and this new set of formulas now represent the coordinates of the center of mass for the rigid body in this case the center of gravity is the same as the center of mass and we also know that mass equals to density row times the volume and if the rigid body has uniform density then density row can also be canceled out from the numerator and the denominator of the equation and now this new set of formulas represent the coordinates of the centroid of volume for this rigid body note that centroid means a geometric Center and in this case the center of gravity the center of mass and the centroid of the volume are all the same point and if the volume has a uniform thickness then we can further reduce the situation from 3D to two dimensional and cancel out the thickness from this equation and now we have coordinates for the centroid of an area and sometimes we can even further reduce the situation by canceling out the constant width of an area and get the coordinates for the centroid of our line the line could be either straight or curved since centroid is the geometric center of a volume or shape if the volume or shape has any excess of symmetry then centroid must be on this axis of symmetry therefore we can easily determine the centroid ation of several common symmetrical shapes such as a cube a circle a rectangle or a straight line this can be used to help find the centroid for other un symmetrical volume or shape so let's look at this example finding the central location for a right triangle with base of B and height of H I will demonstrate this problem using two different approach approaches first of all we need to put this triangle into an XY coordinate system so that we can apply these two equations to find the centroid location since in these two equations X and Y represent the coordinates of an arbitrary Point XY in this triangle let's define a differential element at this location with sides of DX and Dy therefore the area of the differential element da equals to DX * Dy also it is helpful to know the equation for this line which is y = to h / B * X as we learned from linear functions probably in pre-calculus class this is helpful because we will use this as the upper limit when we integrate along the y direction therefore first we calculate for xar the numerator is an integration of x * da integrated along the y axis from zero to the line H over BX and along the xaxis from0 to B and we get 1/3 hp^ squar and we do the same thing with the denominator which is 12 H * B you might notice this is simply the area of the triangle 1 have height times the base that is correct the denominator in this formula is simply the total area therefore xar is calculated to be 2/3 * B and then we do the same thing for the Y Bar the denominator is of course still the total area 1 12 H * B but we need to integrate y da a and eventually we get Y Bar equals to 1/3 H this is the first approach however integration with two variables can sometimes be difficult therefore let's look at a second approach in which we only need to integrate with one variable so we have the same red triangle put in the XY coordinate system but this time we choose a vertical strip of WID DX instead of a little square to be our differential element the height of this strip is determined by the y-coordinate of this point which is H over B * X as we determined earlier from the equation of this line now in these equations as you can see X and Y are replaced by X tutor and Y tutor this is because now my differential element can no longer represent a particle therefore in these two equations xtor and Y Tor represent the coordinates of the centroid location of the differential element which in this case is a rectangle and we know that its centroid is right here at the center and we can tell that X Tor equals to X and da equals to the area of the rectangle which is the height H over B * x * the width which is DX we plug these in into the first equation therefore xar equal to 2/3 B just like we got using the first approach but here as you can see we only need to integrate over one variable X to find Y Bar we do the same thing but use a horizontal strip instead and again we get the same result Y Bar is 1/3 H so it is up to you to decide what is the best method to use as you can see choosing the differential element cleverly can greatly simplify your calculation now we Mark the centroid position on the triangle and as you can see it is located at 1/3 location to either side actually if you recall we used this conclusion before when replacing a linearly distributed loading the one that shapes like a triangle with a concentrated load the centroid locations for common shapes have long been summarized and can be easily found online or in engineering textbook or handbooks for example here is a screenshot of a Wikipedia page on centroids you might find this information very useful