1. EEE340Lecture 041 b). Along path OP 1 P Solution. Using (2-52) of cylindrical a).

2. EEE340Lecture 042 b). The integrals (2-80) and (2-81) are the same mathematically, and they both result in a scalar.

3. EEE340Lecture 043 Example 2-14. where and path C is a portion of circle r =3 in the 1 st quadrant count-clock, as shown in the Figure. Solution 1. Cartesian The integral path satisfies

4. EEE340Lecture 044 Therefore Solution 2. Cylindrical (polar) Advantage: (From (2-52)) Disadvantage: Converting into cylindrical

5. EEE340Lecture 045 Using (2-84) it follows

6. EEE340Lecture 046 Hence where on the integration path. finally

7. EEE340Lecture 047 Surface integrals and where is outward base vector Example 2-15. Given find where S is a cylinder rod of and r=2.

8. EEE340Lecture 048 Solution. 1). Top: Hence 2). Bottom. Same result as 1).

9. EEE340Lecture 049 3). Side wall Finally

10. EEE340Lecture 0410 2-6 Gradient of a Scalar field The gradient of a scalar field V is a vector that represents both the magnitude and the direction of the maximum space rate of increase of V. Physical example: a light ball runs down a mountain (2-86) (2-94)

11. EEE340Lecture 0411

12. EEE340Lecture 0412 Example 2-16. Find for: a). b). Solution. a).

13. EEE340Lecture 0413 at a point (1,1,0), E (1,1,0) = a E E b). (2-77)

14. EEE340Lecture 0414 Practice exercise: Given V = xy + yz + xz, find (a)  V at point P 1 (1,2,3) (b) the directional derivative of  at the same point in the direction toward point P 2 (3,4,4) Solution (a) (b) we have to find the direction vector from P 1 to P 2 first

15. EEE340Lecture 0415 2-7 Divergence of a Vector Field where  v is the volume enclosed by the closed surface S in which P is located. Physical meaning: we may regard the divergence of the vector field at a given point as a measure of how much the field diverges or emanates from that point. the divergence of at a given point P (2-98)