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0:00
Hey friends, welcome to the YouTube channel ALL ABOUT ELECTRONICS. So in the previous video, we learned that what is Del operator and we understood the concept of the Gradient. So similarly, in this video, let us understand the concept of the divergence. So in the previous video, we have seen that this Del operator is the Vector Differential operator. And this is how it can be represented in the different coordinate systems.
0:34
And we have also seen that depending on how the Del operator is used with the scalar and the vector fields, we will get the different operations. For example, we have seen that whenever this Del operator is used with the scalar field, then we will get the gradient. And likewise, when this Del operator is used with the vector fields, then we will get either Divergence or the Curl. So specifically, when we perform the dot product of this Del operator and the vector
1:02
field, then we will get the divergence. So first of all, let us understand what is the significance of this mathematical operation and how it is very useful in the electromagnetic field theory. So if you see this divergence, then it measures the net outwards flux of the vector field from the very small volume or from the infinitesimal small volume around the specific point. Or in other words, we can say that if we have some vector field A, then at any given point
1:30
P, the divergence of this vector field A is the net outward flux per unit volume as this volume shrinks around P. So basically, this divergence measures the net outward flux of the vector field from an infinitesimal small volume around the point. That means this unit volume that we are assuming, it should be infinitesimal small in size. So now, let's see how it can be represented mathematically. So if we have some vector field A, then we know that the flux that is passing through
2:02
the cube can be found using the closed surface integral. And in fact, in the earlier video, we have already seen that. That means if we have some cube, then using the closed surface integral, we can find the total flux that is passing through the closed surface. Now as per the definition of the divergence, this divergence is the flux that is passing per unit volume as this volume shrinks around the point.
2:27
That means if we divide this flux by the unit volume, and as this unit volume tends to zero, then we will get the net flux that is passing from the infinitesimal small volume. And it can also be given as the ∇.A. So here, we know that this del operator in the Cartesian coordinate can be given by this expression. And if we see this vector field A, then it will have the three components in the x, y and the z direction. And as you can see, here each component is the function of x, y and the z.
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But here, for the simplicity, we will simply represent them as the Ax, Ay and the Az. So if we perform the dot product of this del operator and the vector field A, then we will get this (∂Ax / ∂x) + (∂Ay / ∂y) + (∂Az / ∂z) So in this way, using this expression, we can find the divergence of the vector field A in the Cartesian coordinate system. So at any given point, if the divergence is positive, it means that the net flux is going away from the given point.
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On the other end, at any given point, if the divergence is negative, it means that the field is converging into the given point. And if the divergence is zero, it means that the incoming flux at the given point is same as the outgoing flux. So from the electromagnetic perspective, this divergence will help us in understanding the Gauss law of electricity and magnetism. And moreover, it will also help us in understanding the other concepts.
4:02
So that is why it is important to understand this divergence. Now so far, we understood that the net output flux from the infinitesimal small volume can be given by this expression. And that is same as the ∇.A But now, let's also prove that indeed this ∇.A corresponds to the given expression. So to prove that, let's take a small differential volume whose dimensions are dx, dy and dz. And for the given volume, first of all, let's find the R.H.S.
4:35
So basically, here we will find the total flux that is passing through the entire closed surface of the given volume. And we know that, that can be found with the help of the closed surface integral. So basically, here what we will do, here we will find the flux that is passing through the individual surfaces of the volume. And the addition of the individual flux represents the total flux that is passing through the closed surface area.
5:02
So here, first let's find the flux that is passing through the front surface. So to find that, first let's take some reference points in the x direction. So let's say in the x direction, the center of the cube is located at the x. And let's say, the front side of the cube is located at the x + dx/ 2. And likewise, the back side of the cube is located at the x - dx/2. That means if we see this width, then that is equal to dx. Likewise, this width will be equal to dy.
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And this height will be equal to dz. So if we see the volume of this cube, then that is equal to dx times dy times dz. So now, let's find the flux that is passing through the surface, which is located at the location x + dx/ 2. So here, for the individual surface, we know that the total flux that is passing through the surface can be found using the surface integral. And it can be given as this A. dS, where this A represents the vector field and the dS represents the differential surface area.
6:16
So here, for the given surface, this differential surface area will point in the +x direction. So here, for the given surface, if we see the area, then that is equal to this dy times dz. And as I said, it will point in the plus x direction. So whenever we perform the dot product of this dS with the vector field, then after the dot product, we will get only the x component of the vector field. So basically, we will have this dy times dz times the x component of the vector field.
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And here, since we are considering the infinitesimal small volume, so there is no need to evaluate it using this surface integral. And rather, we can write this total flux that is passing through the surface at the x + dx/2 as this Ax times the surface area( dy x dz). And of course, since this front side is located at the x + dx/2, so here we will consider the vector field at the location x + dx/2. Likewise, the flux that is passing through the back side at the location x - dx/2
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is equal to - Ax times this (dy x dz). Because here, for the back surface, the differential surface area will point in the outward direction. That is equal to -x direction. So that is why, here we will have the negative sign. And here, since the surface is located at the location x - dx/2, so here, we will consider the vector field Ax at the location x - dx/2. That means the net flux that is passing through the back surface or this φ'x can be given by this expression.
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So if we combine the total flux that is passing through the front side and the back side, then we will get the total flux that is passing in the x direction. And it can be given by the summation of the two components. That is equal to this φx + φ'x. So now, to further simplify it, let's expand these two terms using the Taylor series expansion. And for that, first let's take the first term. So as per the Taylor series expansion, this term can be written as, this Ax plus this (∂Ax/ ∂x) times dx/2.
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So of course, here we will also have the higher-order terms of the dx, but here, since we are taking the infinitesimal small volume, so the value of the dx is very small. And due to that, here we can neglect the higher-order terms of the dx. So similarly, the second term can be expanded like this. And now, as per our earlier term, if we perform the subtraction of these two terms, then this is how it can be represented.
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So here, these two terms will get cancelled out, and these two terms will get added. And finally, we will get, this ∂Ax / ∂𝑥 times dx. So as I said earlier, instead of writing this Ax as the function of x, y and z, here we are simply writing it as the Ax. And now, if we multiply this term with the surface area, then the net flux in the x direction can be given as, this ∂Ax / ∂𝑥 this dx times dy times dz.
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So in this way, for this given closed surface, we found the net outward flux in the x direction. So similarly, by following the same procedure, we can also find the net outward flux in the y direction. And that will come out as, this ∂Ay / ∂y times this dy times this dx times dz. So for finding the net outward flux in the y direction, we will consider these two surfaces. And similarly, by following the same procedure, we can also find the net outward flux in the z direction.
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So for that, we will consider these two surfaces. So now, the total flux that is coming out of this small volume is the summation of the individual flux. And further if we simplify it, then it can be written like this. So basically, here we found this total flux using the closed surface integral. And now, if we divide by the volume, then we will get the net outward flux per unit volume. So here, the volume of this cube is equal to dx times dy times dz.
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So if we divide this term by the volume, then we will get the net outward flux per unit volume. And here, since we are already considering the infinitesimal small volume, so we can say that, here this Δv tends to zero. That means this expression represents the divergence. As you can see over here, this dx dy dz will get cancelled out in both numerator and denominator. That means here, this term is nothing but the divergence. And that is equal to ∇.A
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So we can say that, indeed this divergence represents the net outward flux per unit volume as the volume shrinks to zero around the measurement point. And in the Cartesian coordinate system, this is how it can be represented. So like we have seen, if we integrate this divergence over the volume, then we will get the total flux that is coming out of the volume. And we know that, the same can also be found using the closed surface integral.
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So this theorem is also known as the divergence theorem. And many times, it is also known as the Gauss theorem. So as you can see, as per this theorem, instead of finding the total flux using the surface integral, using the divergence also, we can find the same flux. And in fact, that is a much easier way for finding the net outward flux. So so far, we have seen the expression of the divergence in the Cartesian coordinate system.
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But similarly, these are the expressions of the divergence in the cylindrical as well as the spherical coordinate system. So as you can see, they are a bit different from the Cartesian coordinate system. So here, we are not going to derive the last two expressions, but we will intuitively understand that how we can write these two expressions. And in that way, you can easily also remember these two expressions.
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So first, let's see the expression of the divergence in the cylindrical coordinate system. So earlier, in the Cartesian coordinate system, when we took the infinitesimal small differential volume, then while finding the net flux in any direction, when we were moving in the specific direction, then in that direction, the area that is seen by the flux remains the same. For example, the area that is seen by the vector field at the x + dx/2 and the x - dx/2 is equal to same.
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And that is equal to dy times dz. So of course, here the value of the flux at the front and the backside will be different based on the vector field. But the area that is seen by the vector field will remain the same. And the same is the case when we are moving in the y and the z direction. But if you see the cylindrical coordinate system, then that is not the case. For example, in the cylindrical coordinate system, when we are moving in the ρ direction,
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then the surface area that is perpendicular to it can be given as this ρ x dφ x dz. where this dz represents this length, and this ρ dφ represents the length of this arc. So that is why the differential surface area can be given as this ρ dφ times dz. So as per this expression, as you further move in the ρ direction, then this area will increase. And that is also evident from the figure. For example, if you see the area on the inner side, then that is smaller than the outer side.
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And naturally, as the area increases, then the amount of flux that is passing through the area will also increase. So that is why, while finding a net flux in the ρ direction, we also need to consider this ρ. And that is why, here this ρ is also appearing in the equation. But let us exactly see why we are getting this term. So now as we have seen in the cylindrical coordinate system, after the Taylor series
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expansion, the net flux in any direction, or in this case, the net flux in the ρ direction will be of the form, this ∂ / ∂ρ times this Aρ times dρ times surface area. And in this case, that is equal to ρ dφ times dz. But as you can see from the expression, here the surface area is also a function of ρ. So while moving in the ρ direction, we also need to consider this ρ. And that is why, this ρ will come inside this bracket.
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That means here, during the partial differentiation, we also need to consider this ρ. While if see this dφ and dz, then that will remain the constant. That means here, this dφ times dz will come on this side. So this will be the net flux that is passing in the ρ direction. Now here, if we divide it by the volume, then we will get the divergence component in the ρ direction. So as you know, in the cylindrical coordinate system, for this infinitesimal small piece,
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its volume can be given as this rho times d5 times dz times dρ. So as you can see over here, these three terms will get cancelled out. And eventually, we will have this 1/ ρ times, this ∂/∂ρ times, this (ρ Aρ). And in this way, we got our first term. So similarly, now let's see this third term. So similarly, now when we move in the z direction, then we will encounter these two surfaces. And as it can be seen from the picture also, while moving in the z direction, the area
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remains the same, or basically, it is independent of the z. And here, this area can be given as this dρ times this ρdφ So once again, as we have seen in the case of the Cartesian coordinate system, as per the Taylor series expansion, the net flux in the z direction will be of the form this ∂/∂z times (Az) times this dz times area. And here, this area is equal to this ρ dφ times dρ. So as it is evident from the expression, here any component of the expression does not contain the variable z.
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That means here, this will be the net flux that is passing in the z direction. And now, if we divide it by the volume, then we will get the component of the divergence in the z direction. So we know that this volume can be given as this ρ dφ times this dz times dρ. So here, all these components will get cancelled out. And we will have this ∂Az /∂z. That is our third term. So similarly, let us also see how we got this second term.
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So now, while finding the net flux in the φ direction, we will encounter these two surfaces. And here, the area of these two surfaces is equal to this dρ times dz. So as you can see, while moving in the ρ direction, the area of both surfaces are remaining the constant. And that is also evident from the shown image. So once again, here the expression of the net flux in the φ direction will be of the form this ∂/∂φ times this Aφ times dφ times surface area. And here, that is equal to this dρ dz.
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So here, since the surface area does not contain any variable of φ, so in the bracket, we will only have this Aφ. And here, this will be the net flux that is passing through the φ direction. And now, if we divide it by the volume, then we will get the φ component of the divergence. So once again, here this differential volume can be given as this ρ dφ times dz times dρ. So here, this dρ, dφ and dz will get cancelled out. And we will have this 1/ρ times this ∂Aφ/∂φ.
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So in this way, intuitively, we can remember the expression of the divergence in the cylindrical coordinate system. So similarly, let's quickly see how to write the expression of the divergence in the spherical coordinate system. So for that, let's take a small volume in the spherical coordinate system. And first of all, let's find out the net flux that is passing through this volume in the r direction.
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So in this radial direction, if you see, then the surface area can be given as this r²sin θ dθ dφ So here, if you see this length, then that is equal to r dθ. And if you see this length, then that is equal to r sinθ dφ. That means the overall surface area is equal to r²sin θ dθ dφ So this will be the one surface. And similarly, we will also have the back surface. So as you can see, this surface area consists of the term r².
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That means as we move radially outwards, then the surface area will increase by the factor of r². And that is why, while writing the expression, we also need to consider this r². So once again, the net flux in the r direction will be of the form, this ∂/∂r times this Ar times this dr times surface area. And that can be given by this expression. So here, since the surface area is also a function of r, so during the partial differentiation, we also need to consider this r².
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While if we see the other terms, then they will remain the constant. That means here, we will have this del divided by ∂/∂r this r² times Ar times dr times this dθ times dφ times sin θ. So this will be the net flux that is passing in the radial direction. And now, if we divide it by the volume, then we will get the radial component of the divergence. So we know that for this differential volume in the spherical coordinate system, its volume can be given as this r² sinθ x dr x dθ x dφ.
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So here, this dr dθ dφ will get cancelled out. Likewise, this sinθ will also get cancelled out. And eventually, we will have this 1/r² times this ∂/∂r times r² times Ar. That is the first term. So similarly, now let's also see how we got this second term. So similarly, now when we move in the θ direction, then we will encounter these two surfaces. And this surface area can be given as this dr times this r sinθ dφ
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That means as we move from the top to bottom side, then in this θ direction, the surface area will increase by this sinθ factor. That means while writing the expression, we also need to consider this sin θ. So once again, the net outward flux in the direction will be of the form, this θ∂/∂θ times this Aθ times dθ. And of course, that will also get multiplied by the surface area. And that is equal to r sinθ times this dr dφ.
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So here, since the surface area also consists of the term sin θ, so while writing this partial differentiation, we also need to consider it. That means here, we will also have the term sin θ, while the rest of the terms will remain the constant. So that is equal to r² x dr x dφ So this will be the net flux that is passing in the θ direction. And now, if we divide it by the volume, then we will get the θ component of the divergence. So here, this volume is equal to r² sinθ times this dr dθ dφ
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So, once again here this dr dθ dφ will get cancel out from the numerator and the denominator. And here, this r will also get cancelled out from the numerator. So eventually, we will have this 1/(rsinθ) times this ∂/∂θ times this Aθ times sin θ. So that is the second component of the divergence. So similarly, now let's see how we got the third term. So now, when we move in the φ direction, then we will encounter these two surfaces.
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And the differential surface area in this direction can be given as dr times this r dθ. So as you can see, it does not contain any term related to the φ. That means while moving in the φ direction, the area of both the surfaces will remain the same. So now, the net flux in the φ direction will be of the form ∂/∂φ times this Aφ times dφ. And of course, it will also get multiplied by the surface area. So that is equal to r dr dθ. So this will be the net flux in the φ direction.
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And now, if we divide it by the volume, then we will get the φ component of the divergence. So we know that this differential volume in the spherical coordinate system is equal to this r² sinθ times dr times dθ times dφ. So once again, here this dr, dθ and dφ will get cancelled out from the numerator and denominator. Moreover, this r from the numerator will also get cancelled out. And effectively, we will have this 1/ (r sinθ) times this ∂Aφ/∂φ So that is our third term.
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And now, when we add all the three components, then we will get the total divergence of the vector field in the spherical coordinate system. So in this way, intuitively, we can remember these two expressions of the divergence in the cylindrical and spherical coordinate system. And from that, using the divergence theorem, we can easily find the net flux that is passing through the given object or the volume.
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And in fact, it is much easier than finding the same flux using the closed surface integral. So here, to understand that, let us take the same example, which we have taken in the earlier video of the line surface and the volume integral. So here, we've been asked to find the flux that is coming out of the closed surface, which is defined by the following limits. So as you can see over here, this vector field is given to us in the spherical coordinate system.
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So earlier, I have asked you to find the flux using the surface integral. But here, let us find the same using the divergence theorem. And then, you can also verify the result in both cases. And for that, first of all, let us find this ∇.A. But before that, let us see the volume that is defined by the following limits. So here, the r is varying from 0 to 4. Likewise, the θ is also varying from 0 to π/2. And here, since the θ is varying from 0 to π/2 only, so we will only have the semi-sphere.
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Provided, the φ is varying from 0 to 2π. But here, the φ is varying from 0 to 3π/2. That means here, we are interested in finding the flux that is passing through the 3 quarters of the semi-sphere. So here, since the φ is 3π/2, so we will not consider this last quarter. And here, the closed area that is defined by this remaining 3 quarters. So basically, here we want to find the net flux that is passing through this closed surface, which is defined by the following limits.
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So here, for that, we will use the divergence theorem. And for finding the same flux using the divergence theorem, first of all, we need to find this ∇.A So in the spherical coordinate system, this ∇.A can be defined by the following expression. So here, one by one, let's find each term. So first, if we see this first term, then that can be given as 1/r² times ∂(r² Ar)/∂r That means here, we will have this 1/r² times this ∂/∂r times r² times the r component of the A.
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And in this case, that is equal to r. That means we will have this 1/r² times this ∂/∂r times this r^3. And we know that the partial differentiation of the r^3 is equal to 3 times r². So we will have this 1/r² times this 3r². And that is equal to 3. That means here, the first component is equal to 3. Similarly, if we see the second component, then that is equal to 1/ (r sin θ) times ∂/∂θ times (Aθ sinθ) So here, the θ component is equal to -3. That means here, we will have this -3 sin θ.
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Now we know that the differentiation of the sin θ is equal to cos θ. That means here, we will have (-3 /sin θ) times cos θ That means here, this will be our second term. And likewise, let us also find the third term. So here, the third term can be given as 1/ (r sin θ) times ∂Aφ/∂φ. So that is equal to ∂/∂φ times the φ component of the vector field. And that is equal to 5 φ. So if we perform the partial differentiation, then we will have this 5/ (r sin θ)
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So in this way, we got all the three components. So if we combine them, then overall this ∇.A can be written as this 3 - 3/r (cos θ / sin θ) + 5 / (r sin θ ) So now, with the help of it, we can easily find the volume integral. So for that, we also need the dv. So we know that in the spherical coordinate system, this dv is equal to r² sin θ dr dθ dφ So if we multiply both of them, then we will get this (3r² sin θ - 3r cos θ + 5r) dr dθ dφ
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So now, we need to integrate this expression over the given range. So that will be equal to the triple integration of this term. So here, what we will do, we will evaluate each term first. So here, the r is varying from 0 to 4, while the θ is varying from 0 to π/2. And likewise, the φ is also varying from 0 to 3π/ 2. So here, if we just consider the first term, then that is equal to 3r² sin θ times dr dθ dφ So further, it can be written as integration 0 to 4, 3r² times dr times integration
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of 0 to π/2, this sin θ times dθ. And similarly, integration of 0 to 3π/ 2 times dφ. That means here, now we can evaluate the individual terms. So we know that the integration of 3r² is equal to r^3 / 3. And here, the limits will be from 0 to 4. Likewise, the integration of sin θ is equal to - cos θ. And here, the limits will be from 0 to π /2. And likewise, the integration of dφ will be equal to φ. And here, the limits are from 0 to 3π/2. So here, this 3 will get cancelled out.
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And we will have this 4^3 times 1 times 3π/2. And that is equal to 96π. So this will be the volume integration of the first term. Similarly, let's also finds the volume integration of the second term. So once again, for the second term, we can write it as integration from 0 to 4 times integration of 0 to π/2 times integration of 0 to 3π/2 times -3r cos θ times dr dθ dφ So here, once again, let us evaluate the individual term.
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So here, the first term will be equal to -3 times integration 0 to 4 times r times dr. Likewise, the second term will be equal to integration 0 to π/2 times cos θ dθ And likewise, the third term will be equal to integration of 0 to 3π/2 times dφ. So we know that the integration of the r is equal to r² /2 . And here, the limits will be from 0 to 4. Likewise, the integration of cos θ is equal to sin θ. And here, once again, the limits are from 0 to π/2.
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And here, for the third term, the integration of dφ is equal to φ. And the limits will be from 0 to 3π/2. That means here, for the third term, we will get 3π/2. So if we further simplify it, then we will have -3 times this 16 divided by 2 times 1 times this 3π/ 2. And that is equal to minus 36π. So this will be our second term. And likewise, let us also find the volume integral of the third term. So for the third term, we can write it as integration of 0 to 4 times integration of
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0 to π/2 times integration of 0 to 3π/2 times 5r times this dθ dφ dr So further, we can write it as integration of 0 to 4 times 5r times dr times integration of 0 to π/2 times dθ. And likewise, the third term will be equal to integration of 0 to 3π/2 times dφ. So further if we simplify it, then we will get this 5 times this (r²/2). And here the limits will be from 0 to 4. Likewise, the integration of the dθ is equal to θ. And here the limits will be from 0 to π/2.
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And likewise, the integration of dφ will be equal to φ. And here the limits will be from 0 to 3π/2. So if we further simplify it, then we will have this (5/2) x 16 x (π/2) x (3π/2) And that is equal to 30π². So that will be our third term. So if we combine all the three terms, then the overall flux can be given as this 96π - 36π + 30π² And that is equal to 60π + 30π². So this will be the overall flux that is passing through the closed surface area which was defined by the given limits.
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So in this way, using the divergence theorem, we can easily find the net flux that is passing through the closed surface. So during this course of the electromagnetic theory, we will use the concept of divergence and this divergence theorem many times. But I hope in this video, you understood the basic concept of divergence and you also understood using the divergence theorem, how we can find the net flux that is passing through the closed surface.
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So if you have any question or suggestion, then do let me know here in the comment section below. If you like this video, hit the like button and subscribe to the channel for more such videos.
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