To understand what the divergence is measuring, look at the following figure that represents a small cube in a vector field.
If the cube is small enough, the vector field will not change much over the cube so a representative vector is drawn on each of the faces of the cube. First let's see if more of whatever the vector field represents is leaving the cube than is entering the cube. Let the vector field is given by F(x,y,z) = f(x,y,z)i + g(x,y,z)j + h(x,y,z)k. Along each face of the cube, the only part of F that is entering or exiting the cube is the component of F that is orthogonal to the face. For the faces parallel to the yz-plane, the net F exiting the cube is the difference in the i component of F from the left face to the right face (the two green vectors) since the other components are parallel to the face. If the distance is small, this can be approximated by f x dx. Since the green vectors were only representatives, every point on the face has an equivalent vector which you can approximate by the same green vectors. So the net F crossing in the x-direction is f x dx ( dy dz ) where dy dz is the area of the face. A similar argument can be made for the other four faces. Finally, the net F exiting the cube is
| ( f x + g y + h z ) dx dy dz = DIV F dV |
Another way you can measure the net F crossing the surface is to sum up all the components of F that leave the surface on each face ( if the term is negative, that means F is entering the face.) Again, it is only the orthogonal component of F that is leaving the face. Let N i be the outward unit normal to the i th face. The amount of F leaving on that face is F • N i dA i , where dA i is the area of the face. The factor N i dA i is referred to as dS. If you partition a solid with these little cubes, the amount of F exiting one cube face is entering the face of the cube adjacent to it. Therefore, they will cancel each other out. This will happen to all the faces of the cubes except those that border on the boundary of the solid. This means that if you want the net F exiting the solid, you can sum (integrate) up all the F •dS terms on the surface of the solid. Equivalently, because
you can also get the net F leaving the solid by summing (integrating) DIV F dV over the solid. This is called the Divergence Theorem. To use the Divergence Theorem, the vector field F must have continuous first partial derivatives, the bounded surface must be closed and have an outward normal. Then
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The surface integral on the left is referred to a Flux integral.
1. Use the Divergence integral to find the flux integral of the given field F across the given surface.
The figure above shows a small rectangular closed curve parallel to the yz-plane. Again, if the dimensions of the rectangle is small enough, the vector field, F(x,y,z) = f(x,y,z)i + g(x,y,z)j + h(x,y,z)k, is "almost constant" on each side of the rectangle and can be represented by the field at the center of the side. Instead of looking at the net F crossing the curve, you want to look at how F is trying to rotate the figure (think of the vector field as some kind of force field for this example.) In this case, it is only the components of F that are parallel to the path that is going to try and rotate the figure. Look at the blue vectors first. If the component of F in the y-direction at the top of the rectangle is bigger, however slightly, than the same component of the bottom vector, then the figure will try and rotate clockwise about a line parallel to the x-axis. The change of this component can be approximated by g z (x,y,z) dz. Since the blue vectors represent all the vectors along the horizontal side which has length dy, the total change for the whole side is g z (x,y,z) dz dy. Writing this rotational effect as a vector is g z (x,y,z) dz ( - 1 ). (Note: the minus is because the rotation is clockwise which is a negative rotation.) Now look at the red vectors. It is the component of F that is parallel to the z-axis that is parallel to the path. If this component is bigger on the right than on the left side, the rectangle will rotate counterclockwise about a line parallel to the x-axis. Using partials to approximate the change and writing the total effect as a vector will give you h y (x,y,z) dy dz. The net rotational effect will be the sum of these or ( h y (x,y,z) - g z (x,y,z) ) dy dz. This same analysis can be done for small rectangles parallel to the other two coordinate planes. The result you will get is summarized in the table below.
| ( h y (x,y,z) - g z (x,y,z) ) dy dz | Parallel to the yz-plane |
| ( f z (x,y,z) - h x (x,y,z) ) dx dz | Parallel to the xz-plane |
| ( g x (x,y,z) - f y (x,y,z) ) dx dy | Parallel to the xy-plane |
| CURL F • dS | where dS = dy dz i, dx dz j or dx dy k |
Another method of calculating this effect is to sum up the components of the vector field that are parallel to the curve directly. For the top blue vector, this would be F •( - j ). The direction - j was chosen instead of j so that a positive answer means the rotation is counterclockwise. Again, this vector represents all the vectors along the upper side, so the total for this side is F •( - j ) dy. For the edge on the right it would be F • k dz. The other two sides are included in the table below written as a vector that denotes rotation about a line parallel to the x-axis in a counterclockwise direction.
| F •( - j ) dy | top edge |
| F • k dz | right edge |
| F • j dy | bottom edge |
| F •( - k ) dy | left edge |
| F • dr | where dr = - j dy, k dz, j dy or - k dz |
You can do a similar analysis for small rectangles parallel to the other two coordinate planes and the equations given in the bottom rows of the two tables will hold for appropriate values of dS and dr.
Now look at any surface in three space. Partition the surface into small rectangles. If two rectangles have a common side then the F • dr quantity will be equal in magnitude but opposite in sign. To see this put two rectangles parallel to the yz-plane side by side horizontally. The right side of one rectangle is the left side of the other. The vector field is the same on both sides by dr = k for the right edge and - k for the left edge. The only rectangles whose F • dr values do not cancel are those rectangles the border the edge of the surface. This means that the sum of all the rectangles is equivalent to the sum (integral) of F • dr along the curve that represents the boundary. However, each F • dr around the small rectangle is also equal to CURL F • dS for that rectangle. So you get the theorem
Stoke's Theorem: Let S be a surface bounded by a
piecewise closed
smooth curve. Choose the
unit normal, N, to the surface so that the curve is counterclockwise with
respect to
N. If the
vector field F has continuous partial derivatives on an open region containing
S and C
then
where dS = N dS |
The line integral above is called the Circulation of F around C.
2. Use the line integral to find the circulation of F around the C in the counterclockwise direction.
Now look at the surface integrals that showed up in both the Divergence and Stoke's Theorems. Surface integrals are calculated by projecting the surface onto one of the coordinate planes. The projected region is partitioned into small rectangles and a double integral is set up. The limits of integration will be the same limit you would use if you were calculating the surface area. Remember from your work on surface areas, the relationship between the area on the surface and the area of the small rectangle in the coordinate plane that is its projection is the magnitude of the secant of the angle the normal to the surface makes with the axis orthogonal to the coordinate plane. ( If you project it to the xy-plane, this will be the secant of gamma. ) This secant can be obtained from the big gradient to the surface. The table below lists these relationships.
| Let a surface be described by f(x,y,z) = constant | |
 |
Projecting to the xy-plane and using the normal with positive z-direction |
 |
Projecting to the yz-plane and using the normal with positive x-direction |
 |
Projecting to the xz-plane and using the normal with positive y-direction |
From your work on tangent planes, you
know that
 . So the vector form using dS = N dS is | |
 |
Projecting to the xy-plane and using the normal with positive z-direction |
 |
Projecting to the yz-plane and using the normal with positive x-direction |
 |
Projecting to the xz-plane and using the normal with positive y-direction |
3. Use the surface integral to find the flux integral of the given field F across the given surface.
4. Use the surface integral to find the circulation of F around the C in the counterclockwise direction.
5. You do not need a closed surface to calculate the flux across a surface. Find the flux of F across the surface S using the normal pointing away from the origin.
© 2003: Roxanne M. Byrne
