Example.1. Calculate the flux of the vector field over the surface of a unit cube whose edges are parallel to the axes and one of the corners is at the origin. Use this result to illustrate the divergence theorem.
- The geometry of the cube along with the direction of surface normalsare shown in the figure. Consider the base of the cube which is the plane z=0. On this face
- Since the normal is along the -z direction flux from this face is zero. Similarly, the flux from the other two faces which meet at the origin are also zero. Consider the top face where z=1.
- On this face The normal is in the +z direction, so that the flux is Likewise, the flux from the other two faces are also ½ each. The total flux, therefore, is 3/2. The divergence of the field
- The volume integral of the divergence is
- By symmetry, this is 3 times 1/2. Thus the volume integral of the divergence is 3/2.
Example.2. Calculate the flux of the vector field over the surface of a unit sphere. Use this result to illustrate the divergence theorem. (Use spherical coordinates).
- Divergence of the field is Thus the volume integral of the divergence over the surface of a unit sphere is just
- To calculate the surface integral we note that the normal on the surface of the sphere is along the radial direction and is given by where R=1 is the radius of the sphere. Thus
- Since the surface element on the sphere is we have , substituting (in spherical polar)
- The first term gives zero because of vanishing of the integral over φ. We are left with
Example.3. Calculate the flux of the vector over the surface of a right circular cylinder of radius R bounded by the surfaces z=0 and z=h. Calculate it directly as well as by use of the divergence theorem.
- Let the base of the cylinder be at z=0 and the top at z=h. The origin is at the centre of the base. The cylinder has three surfaces. For the bottom surface, the direction of the normal is along and on this surface z=0.
- The surface integral for this surface is For the top surface, the normal is along the surface integral is For the curved surface the direction of the normal is outward radial direction in the x-y plane which is so that the surface integral is
- The integral is done in the cylindrical coordinates by polar substitution The surface element is Thus the integral become The angle integral in both cases gives zero. Thus the total flux is only contributed by the top surface and is The angle integral in both cases gives zero. Thus the total flux is only contributed by the top surface and is This can also be seen by the divergence theorem. The volume integral is
Example.4. Calculate the flux of the position vector through a torus of inner radius a and outer radius b. Use the result to illustrate divergence theorem. (* This is a hard problem).
- Geometrically a torus is obtained by taking a circle, say in the x-z plane and rotating it about the z-axis to obtain a solid of revolution. Let us define the mean radius of the torus to be and the radius of the circle which is being revolved about the z-axis to be
- The position vector of an arbitrary point on the torus is defined as follows:
- Consider the coordinate of an arbitrary point on the circle which is in the x-z plane. Let the position of the point make an angle φ with the x-axis. The coordinate of this point is When the circle is rotated about the z-axis by an angle θ, the z coordinate does not change.
- However, the x and y coordinates change and become
- Thus an arbitrary point on the torus can be parameterized by given by the above expressions. A surface element on the torus is then obtained by the area formed by an arc obtained by incrementing and the arc formed by incrementing
- The area element is therefore given by the cross product (The unit vector on the surface is directed along the direction of the cross product.). The partial derivatives are given by
- Thus (the order of the cross product determines the outward normal).
- Thus , substituting
- Since the integrand is independent of θ the integral over it gives 2π. The integral over the remaining angle is straightforward. We can simplify the integrand as follows :
- The second and the third term in the integral vanish, the remaining two terms give which makes the total contribution to the surface integral as
- However, the problem is straightforward if we apply the divergence theorem. The divergence of the position vector is 3. Thus by divergence theorem, the surface integral is 3 times the volume of the toroid.
- The volume of the toroid is rather easy to calculate if we note that if we cut it along a section, the toroid becomes a cylinder of radius r and length 2πR. Thus the volume of the toroid is Thus 3 times the volume is consistent with our direct evaluation of the surface integral.
Example.5. Calculate the flux of the vector field over the surface defined by
- The surface is sketched below. Since the region of interest is Notice that the divergence theorem is not directly applicable because the surface is not closed.
- However, one can close the surface by adding a cap to the surface at z=3. We will calculate the flux by applying the divergence theorem to this closed surface and then explicitly subtract the surface integral over the cap.
- The divergence of the field is given by
- Thus the surface integral over the closed surface is thus given by To evaluate this, consider a disk lying between z and z+dz. The circular disk of width dz has a volume
- This gives the surface integral to be 4π. We have to now subtract from this the surface integral over the cap that was added by us, which is directed along direction. On this surface z=3 so that the radius of the disk is 1. This integral can be easily calculated and shown to have value zero. Thus the required surface integral has a value 4π.
In the following questions(Q1-Q5) calculate the flux both by direct integration and also by application of the divergence theorem.
Q.1. Calculate the flux of the field over the surface of a right circular cylinder of radius R and height h in the first octant, i.e. in the region (x>0, y>0, z>0).
- For the curved surface of the cylinder, the unit vector is which gives
- Parameterize Since we are confined to the first octant
- The flux through the slant surface is The top and the bottom caps are in the the contribution from these two give zero by symmetry. T
- here are two more surfaces if we consider the first octant, they are the positive x-z plane and the positive y-z plane., the normal to the former being in the direction of that for the latter is along
- The flux from the former is while that from the latter is
- Adding up all the contributions, the total flux from the closed surface is zero. This is consistent with the fact that the divergence of the field is zero.
Q.2. Evaluate the surface integral of the vector field over the surface of a unit cube with the origin being at one of the corners.
- There are six faces. For the face at x=0, since the surface is directed along the surface integral is The face at x=1 gives +1/2. The faces at y=0 and that at z=0 gives zero because the field is proportional to y and z respectively.
- The contribution to flux from y=1 is 1 and that from z=1 is 3/2. Adding, the flux is 5/2 units. This can also be done by the divergence theorem. Divergence of the field is 2x+3y, so that the volume integral is
Q.3. Calculate the flux of over the surface of a sphere of radius R with its centre at the origin.
- The divergence of the given vector field is Thus, by divergence theorem, the flux is
- We can show this result by direct integration. The unit normal on the surface of the sphere is given by that the flux is This integral can be conveniently evaluated in a spherical polar coordinates with The surface element on the sphere is
- By symmetry, the flux can be seen to be . (Note that we decided to do the integral involving z4 rather than x4 or y4 because the azimuthal integral gives 2π in this case. The integral is easy to perform with the substitution which gives the flux to be
Q.4. Calculate the flux of through the surface defined by a cone
- The divergence of the field is 3. The flux, therefore, is 3 times the volume of the cone which is The flux is thus π. The direct calculation of the flux involves two surfaces, the slant surface and the cap, as shown in the figure.
- The cap is in the xy plane and has an outward normal (because on the cap z=1 and the cap is a disk of unit radius). Thus it remains to be shown that the flux from the slanted surface vanishes. At any height z, the section parallel to the cap is a circle of radius z. Since, the height and the radius of the cap are 1 each, the semi angle of the cone is 450.
- Thus the normal to the slanted surface has a component along the z direction and in the x-y plane. The component in the xy plane can be parameterized by the azimuthal angle φ and we can write The area element can be written as appears because the length element is along the slant.
- Thus the contribution from the slanted surface is this integral can be evaluated and shown to be zero.
Q.5. Evaluate the flux through an open cone for the field
- This problem is to be attempted similar to the problem 5 of the tutorial, i.e., by closing the cap and subtracting the contribution due to the cap. The divergence being 3, the flux from the closed cone is 3 times the volume of the cone which gives 8π.
- The contribution from the top face (which is a disk of radius 2) is the net flux is zero. (You can also try to get this result directly as done in problem 4, where we showed that the flux from the curved surface is zero).