 Okay now do one thing, take a rod, this is a uniform mass distribution rod, mass is in, length is L, okay, find out its centre of mass using this formula, we know that centre of mass will be L by 2, we know that, prove it, using that integral formula, take a dm mass, it will have dx length at a distance x away, what an integral physics initially will face trouble, so you can't ignore it, you have to get through that, slowly as slowly you will get a hang of it. See you have to take a small mass dm over here, see mass is spread, okay, so let's say the length of dm is dx and it is at a distance of x away. So let's say the length of dx, the entire length will have, so this is my dm, so I have got dm in terms of what you have to do regular thing, let's say only come once and the different scenario will have different way of doing problems and this you have to do again and again, so better do it in the next scenario, okay, and yes, in the school exams they will not ask you to integrate and find out the centre of mass locations, but they expect you to do in all the public exams, so you should know this. And anyway if you know only things which everybody, so 50 lakh people write the exam, your rhyme will also be 5 such lakhs, right, because you have not learnt something which others don't know, so why your rhyme will be better than them, isn't it? So you should pay attention here, you should know something which others don't know and there is nothing very complicated here. So interior of h dm will become what? Integral of x m by l into dx divided by m only, so this is my x centre of mass, my limits of interior will be what? From 0 to n, x is going from here to here, it's the mass that are distributed starting from here to there, okay. Now try finding out the answer quickly, m and m goes away, 1 by l comes out of it, integral is constant, 0 to l x dx, x dx integral is what? x square by 2, 0 to l are the limits, so you get l by 2, l square by 2 will come and 1 l will get cancelled, so l by 2. But suppose you have a rod length l, okay, mass per unit length is given by this, x by l kg per meter, you need to find centre of mass of this rod now, will it be centre of the rod? No, mass per unit length is increasing, so where which side it will be? Okay, so your x starts from here, left hand side, but where it will be? Same way, exact centre, answer has to be more than l by u, otherwise the answer will come. Should I do it? Lambda into dx is dm now, mass per unit length is directly given, in earlier case you found out mass per unit length m by l, because it was uniformly distributed. Mass per unit length is kg per meter, that into length is the mass, okay, imagine a dx element over here at a distance x 7, what will be the mass of this dx? dm will be what? Lambda into t x or not? x dx by n and out here, mass per unit length into the small length will be equal to the small mass, okay, it's a usual way. So integral of will become integral of, so the limits will go from 0 to n divided by, what is the answer? 2 by 3 mass location, it will not be like simple state path, okay, can you tell, you can say this is my y coordinate of mass. Try doing, in this case, 8 times of theta, so if this is theta, okay, your dm is the mass, find out everything, so you have cancelled out like this, find theta, d theta, limits will be what? Okay, and theta, it is negative, so this will be equal to minus of 2 r of mass, so use that to find out the center of mass for the mass is distributed on the a, so you have to find mass into small r into width, which is dm, pi into capital pi into small r, this is my dm, minus 2 small r by pi, this is the center of 0 to small r, integral of the r of mass, and you are integrated. Not cancelled, no, it will not, pi r square integral 0 to r, r square d r, so when you put limit 0 to, sorry, this is capital r of mass, okay. See why we are doing this, nobody is going to America, okay, this is not part of theta.