is called a surface integral. Because the surface is an area, surface integral is an area integral.

Note that in the surface intergral, we mentioned a function that applies on the surface of the solid. To have a visualization of this concept, consider a tank with water entering and leaving it. The function that would apply on the surface of this tank could be the velocity of the water that enters it and the velocity of the water that leaves it. (The places where the water enters and leaves the tanks are surfaces of the tank, where holes are cut through for the water to enter and leave. The other portions of the tank would not be open to the water; thus, no function would apply on these portions.) The surface integral could then be applied to the velocity functions upon entry and exit of the water.

Now, also, note in the figure the differential volume dV. This differential volume is not on the surface of the solid but is inside it. Assume a function exists that applies inside the volume. As in the case of the surface integral, this function can be integrated over the entire volume and, to do this, the symbol L is used. Let the

function bef(V) = V + 3V and the result of the integration be Vs. Thus,

fV(V)dV = fV(V + 3 V)dV is called a volume integral.

An example of a function that applies inside the volume could be anything that transpires inside the volume. Again, consider the tank above, where this time, there is a chemical reaction occurring inside it. This chemical reaction can be expressed as a function; thus, it can be used as the integrand of the volume integral.


A vector is a quantity that has magnitude, direction, and orientation. In^this book, we will use the arrow on the roof of a letter to signify a vector. Therefore, let S represent any vector. In the Cartesian coordinate system of xyz, this vector may be decomposed into its components as follows:

Sj, S2, and S3 are the scalar components of S in the x, y, and z directions, respectively, and n1, n2, and n3 and are the unit vectors in the x, y, and z directions, respectively.

Refer to Figure 1. As shown, dA has the unit vector n on its surface. The component of dA on the x-y plane is dxdy. As shown in the figure, the unit vector normal to dxdy is n3, which is in the direction of the z axis. In vector calculus, the component of dA can be obtained through the scalar product of n and n3. This product is designated by n • n3. In other words, dxdy = n • n3 dA (57)

Another vector that we need to discuss is the nabla, V. This vector may be decomposed in the cartesian coordinates as follows:

The dot (or scalar) product is actually the product of the magnitude of two vectors and the cosine between them. The cosine of 90° is equal to zero, so the dot product of vectors perpendicular (orthogonal) to each is equal to zero. The unit vectors in the Cartesian coordinates are in an orthogonal axes; therefore, their "mixed dot products" are equal to zero. Thus, n, • n2 = 0; n, • n3 = 0; and n2 • n3 = 0 (59)

Now, the cosine of 0° is equal to one. Thus, the dot product of a unit vector with itself is equal to one. In other words, n, • n2 = 1; n2 • n2 = 1; and n3 • n3 = 1 (60)

Usi^g these new-found formulas, we evaluate the dot product of nabla and the vector S. This is written as follows:

Simplifying the above equation will produce

Gauss-Green Divergence Theorem

This theorem converts an area integral to a volume integral and vice versa. Let S be any vector and n be the unit vector normal to area dA, where A is the area surrounding the domain of volume y. Refer to Figure 1. Now^ form the volume integral JV(dS3/dz) dV, where S3 is the scalar component of S on the z axis. Thus,

J d-dV = J d3dxdydz = J jj 2d- dz jdxdy = J jj 2d- dzn • n3] dA

where n3 is the unit vector in the positive z direction.

JA{J/ itdz^n ' "3]dA = L {53(x, y, Z2)}[n • n] dA2

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