Читать книгу RF/Microwave Engineering and Applications in Energy Systems - Abdullah Eroglu - Страница 44

Let's calculate the right‐hand side of Eq. (1.67).

(1.69)

In (1.69), surface areas are defined by

(1.70a)

(1.70b)

(1.70c)

To be able to find the differential surface area vector for surface 4, we need to determine the normal vector for that surface. The normal vector can be found using the equation

Figure 1.17 Geometry of Example 1.4.

(1.71)

In (1.71), f is the equation that defines the surface. The equation that defines the surface is given as

(1.72)

Constants A, B, and C are obtained from the geometry at the intercept points which are (a,0,0), (0,b,0), and (0,0,c). When intercept point a is on the axis with y = 0, and z = 0 are substituted into (1.72), we obtain

(1.73)

Similarly, from intercept point (0,b,0)

(1.74)

and from intercept point (0,0,c)

(1.75)

Then, function f is defined by

(1.76)

Hence, from (1.71) and (1.76)

(1.77)

We can then calculate the surface area as

(1.78)

where

(1.79)

α in (1.79) is the angle between the xy plane and surface 4. Substituting (1.79) into (1.78) gives

(1.80)

Then, from (1.69)

(1.81)

In (1.81), the first three terms are zero, hence (1.81) reduces to

or

(1.82)

From (1.76)

(1.83)

Hence, (1.82) can be written as

(1.84)

(1.84) leads to

(1.85)

The left‐hand side of the equation now can be found as

(1.86)