 This is Dr. Rupali Sherke working as an associate professor at Wolchen Institute of Technology in the electronic department. In this video, we are going to discuss with the problem based on the Gauss-Divergence Theorem. In the previous video, we are discussed with the Gauss-Divergence Theorem. In this video, we will see the problem related to that. At the end of learning outcomes, at the end of this video, students are able to solve or evaluate both sides of the divergence theorem. For any vector, for any vector, what is a divergence statement? It is a closed integral over the surface for any vector d, a bar ds is equal to the integral over the volume del dot a bar dv which states that the integral of normal component of any vector over a closed surface is equal to the integral of the divergence of this vector through the volume. Now, for this, let us take a problem. Problem statement is evaluate both sides of the divergence theorem for a volume of a cube. A cube is given whose sides is 2 meter and centered at the origin such that the edges are parallel to the axis. That is evaluate it. The same statement I had written over here. Evaluate both sides of the divergence theorem of the volume with the cube 2 meter of side centered at origin with the edges parallel to the a x bar and for that it is given a d bar is given to you, d bar is 10 x cube by 3 that is a x bar that is nothing but a vector is given to you. Now, we are already discussed this statement for this, first for this problem statement we will try to draw the diagram what they are given it is given a cube, a cube is given a cube is given whose sides is 2 meter and it is placed exactly at the origin that is 4 we will place an axis from the which is passing through the origin. Now, we will consider 3 axis x axis, y axis and z axis. If you see that it is the it as the side is 2 meter, 2 meter side it is changing from minus 1 to 1 in a y direction. In this direction it is changing from plus 1 to minus 1 in a x direction for the z also it will vary from minus 1 to 1 means the integral values are changing from minus 1 to plus 1. Let us derive this thing. Now, as the statement from the theorem close integral of d ds is equal to rho v del dot d bar dv. We will consider the LHS side LHS will be integral of close d ds is equal to now as it is a close we have to see all the surfaces front surface, back surface, left surface, right surface, right side as well as the top and bottom therefore, it will be a close integral will be front plus back plus right left top and bottom. Now, we will find out the d is given to us, d bar is given to us we are supposed to find out the ds for the front side, ds for the back side, ds for the left, right, top and bottom. Now, for the first surface ds for the front side, ds for the front side this is a front surface which we are seeing this is the front surface what we are observing. This surface this surface the front surface we are observing in a which direction x direction that is a positive x direction and what is varying in here the variation the variation of this here two surfaces what is varying here z is varying and here y is varying here z is varying and y is varying therefore, the area of this is given by dy change, change in dy and change in dz in a which direction we are observing in a front direction that is the ax bar similarly, the back side the surface will be same dy into dz, but only the direction will change it will be in a minus ax direction it will be in a minus ax direction. Similarly, for the right side this surface the next surface we are observing this surface this surface how we are observing we are observing this in a y direction that is positive a y direction. Now, what is varying in this in this this variation this variation is a x variation and this variation is a z variation z and x means z x dz is varying and dx is varying that is how it will be this right side it will be dx dz minus sorry a y and for the left side it will be dx dz only, but only minus a y direction will change. Similarly, for the top surface when you observe the top surface this is a top surface variation for the top surface variation what it is varying dy is varying and dx is varying and we are observing in a a z direction a z direction that is why the top surface will be dx dy a z and bottom it will be dx dy same only direction will be minus a z. Now, we will substitute all these values over the in this equation. Now, we know the one more relation that is a dot product property self dot product is always equal to 1 while with other it is always equal to 0 if you see that a x dot a x is always equal to 1 a x with a y or a x with a z is equal to 0. Now, our given equation of the D is 10 x cube by 3 a x bar it is 10 x cube by 3 a x bar when you take a dot product of a x a x it will be 1 when it is with a y and z it will be 0. Therefore, when you integrate it with respect to LHS will remain only with the front plus back because here front is a x and back is also minus x with this it will be 1, but with respect to left right it will be y and z therefore, when the dot product with the y and z it will be 0. Therefore, left right top and bottom integrals will be equal to 0 from this equation this term this terms till here it will be equal to 0 only you remain with the LHS will be equal to DDS. Now, we will consider this and solve the equation. Now, when we consider DDS left and it will be what is a given DDS is a 10 x cube by 3 10 x cube by 3 it is a x and what is a DS for that front side it will be a for the front side we are already seen dy dz a x that will be dy dz a x again this a x into a x will be equal to 1 and only you remain with the s 10 x cube by 3 dy dz when you integrate this will be a constant term for this integral and only you will integrate with respect to dy and with respect to dz what is the value of y and z already minus 1 to 1 after integral this will be a 40 by 3. Similarly, for the back side DDS this will be same 10 x cube by 3 given it will be a x it will dy dz into minus a x this will be a minus a x this will be a when you integral again this value will be 30 80 the total front and back LHS side will be equal to 80 by 3 80 by 3. Now, let us see RHS side the RHS side is a integral over the volume del dot d bar into dv first we will solve del dot d bar del dot d bar is equal to dou by dou x dou by dou y dou by dou z but if you see that only x component is given the other y and z will be equal to 0. So, we will differentiate only with respect the d bar with respect to x dou by dou x the value whatever it is given it is after differentiating it will be 30 x square by 2 integral this over the volume that will be 10 x square we will consider over here and what will be dv now dv will be integral of the v 10 x square dx dy and dz volume is a product of dx dy and dz it will be integral 10 will be constant it will be integral x square dx integral dy integral dz again it is varying from minus 1 to 1 minus 1 to 1 minus 1 to 1 after solving this you will get it as a 80 by 3 this is a LHS is equal to what we have proved LHS is equal to our LHS value was 80 by 3 80 by 3 again after integrating this is also we are getting is a 80 by 3 this is what the both side of the divergence theorems are being verified this is how we are evaluated the both side of the divergence theorem in this problem statement these are the references thank you