 Now, we will try to quickly solve this centroid problems and the first problem is to find out the centroid of this area using direct integration. So, just look at the problem carefully remember this there is a circular sector at the middle that is defined by the angle alpha and alpha. So, that is a circular sector and then you have two triangular parts on the right hand side and on the left hand side. Remember it can be done in both approaches one is the direct integration the other one is of course, using the you know composite area concept. So, you should be able to verify your answer using the both approaches. So, first try to just think that in the direct integration method if I want to adopt then how do you choose the strip in this case? Can you give me the answer let us say for alpha equals to 30 degree alpha equals to 30 degree can you please give me the answer because the expression that we are going to get could be a bit lengthy. So, if you can give the answer for alpha equals to 30 degree alpha equals to pi by 6. I do not think we are getting the correct result so far. Someone said y equals to 0.48 r that is I think close answer for alpha equals to 30 degree you should get an answer roughly 0.49 r or let us say approximately 0.5 r. So, you see the way the strip is considered we have taken a vertical strip. So, that vertical strip we have defined a theta. So, theta is measured from the y axis. So, ultimately if you look at a d theta therefore, what I have that x equals to. So, this strip I want to integrate from this strip is running along the x axis. So, ultimately what has been done here is that if we look at it that x should be equals to r cosine theta sorry y should be r cosine theta and x should be r sine theta. Now, what is this d x that d x remember that d x will be r d theta cosine theta. So, the d x is chosen very carefully in terms of theta. So, the my integration limit will always be from negative alpha to positive alpha or since we have a symmetry we can also go from 0 to alpha and just multiplied by 2. So, the only trick in this problem is that to carefully calculate d x in terms of theta. So, once that is done we can clearly see here that the area can be calculated which is integral of d a. So, what is my d a now d a is y multiplied by d x. So, y is r cosine theta right y is r cosine theta and d x is also r cosine theta d theta. So, therefore, I have r square cosine square theta d theta. So, in that way I could get the area which is 2 times 0 to alpha and then I have y d x substituted in the forms of theta. So, I get the area. So, now I have the first moment about the x axis. So, first moment about the x axis is nothing, but this y element has to be calculated that is nothing, but r cosine theta divide by 2 the centroid is r cosine theta divide by 2 you have the d a. So, finally, we get the first moment as this and then we just say y bar a equals to this first moment we are equating. So, the final answer will look like this. If you substitute alpha equals to pi by 6 that is 30 degree then we are going to get a solution of 0.5 r roughly 0.5 r approximately 0.5 half r r over 2. So, now you can verify also your answer by the composite area. So, if you want to do it in a composite section form see basically I have 2 triangles and then I have the sector here. So, I take this as a circular sector and then I have 2 triangular areas. So, remember I know the solution for the total area now. So, as you can see for the sector the solution is alpha r square that is the area and then for the triangular part you can also get the area in terms of r and alpha. So, which is written here half r sin alpha multiplied by r cosine alpha multiplied by 2. So, you have 2 areas here. So, I get the total a and then you take the first moment of the individual area by knowing their centroid. So, now I have the expression for y 1 a 1 plus y 2 a 2 plus y 3 a 3. So, what we can do we already know the centroid of the circular sector from the vertex. So, that has already been done. So, I have used that I know the area. So, I have this portion right here that is coming from the circular sector and then also from the triangle I have this part. So, 2 times of the centroid multiplied by the area. So, y 1 a 1 let us say y 2 a 2. So, this part is from the triangle this part is from the circular sector. So, these are giving me the first moment of area about the x axis. Finally, what we have basically this divide by that. So, you can clearly see we are getting this result back. So, this result is what we also got from the direct integration. Once you substitute alpha equals to pi over 6 the answer should be. So, that is the answer. So, roughly r over 2 will be the answer. So, we can verify this problem using two different approaches. However, the main reason to give this problem is it is a bit difficult for the students to understand how this strip will be chosen such that you can you know do the integration ok.