CHAPTER 5 VECTOR INTEGRATION INTEGRAL THEOREMS AND So that we have 4  2 4  320 Example 5.21 Verify the divergence theorem for 4xiˆ 2 taken over the region   F y ˆj z 2 2kˆ bounded by x2+y2=4, z=0 and z=3 Z S Volume Integral FdV  r : z  3 2 S3 Y O 2 z0 1 : z  0 X S  84 The surface S of the cylinder consists of a base S1(z=0), the top S2(z=3) and the convex portion S3(x2+y2=4). Then Surface Integral S S1 S2 F ndS F ndS1 F ndS2    3 F ndS  4xiˆ 2 0, F  y F nˆ  so that ˆj;  2 z 0; On S1  nˆ  1 F nˆdS  S1 z  3;   4xiˆ 2  9kˆ;  9, On S2 nˆ kˆ; F y and F nˆ so ˆj  F  dS2 2 S2 2 S2
CHAPTER 5 VECTOR INTEGRATION AND INTEGRAL THEOREMS So that we have 4  2 4  320 Example 5.21 Verify the divergence theorem for F  4xiˆ 2 y 2 ˆj  z 2kˆ taken over the region bounded by x2+y2=4, z=0 and z=3 Z Volume Integral  FdV r O S : z  3 2 S3 Y 2 z0 X S 1 : z  0  84 The surface S of the cylinder consists of a base S1(z=0), the top S2(z=3) and the convex portion S3(x2+y2=4). Then Surface Integral F  ndS F  ndS1 F  ndS2 F  ndS S S1 S2 3 On S1 z  0; nˆ  F  4xiˆ 2 y 2 ˆj; F  nˆ  0, so that F  nˆdS S1 1 On S2 z  3; nˆ  kˆ; F  4xiˆ 2 y 2 ˆj  9kˆ; and F  nˆ  9, so F  S2 2 dS2 S2