Читать книгу RF/Microwave Engineering and Applications in Energy Systems - Abdullah Eroglu - Страница 47

We first calculate the left‐hand side of Eq. (1.87) as

Figure 1.18 Illustration of Stokes' theorem.

Figure 1.19 Geometry of Example 1.5.

which leads to

The differential surface area is found from

Then

Now, we calculate the right‐hand side of the equation for each segment as

1 x = 0, z = 0, ,

2 x = 0, y = 1, ,

3 x = 0, z = 1, ,

4 x = 0, y = 0, ,

Hence

As a result, it is confirmed that

for Example 1.5.