By B. Hague D.SC., PH.D., F.C.G.I. (auth.)

ISBN-10: 0412207303

ISBN-13: 9780412207303

ISBN-10: 9400958412

ISBN-13: 9789400958418

The important alterations that i've got made in getting ready this revised version of the e-book are the next. (i) Carefuily chosen labored and unworked examples were additional to 6 of the chapters. those examples were taken from category and measure exam papers set during this college and i'm thankful to the collage court docket for permission to exploit them. (ii) a few extra topic at the geometrieaI program of veetors has been included in bankruptcy 1. (iii) Chapters four and five were mixed into one bankruptcy, a few fabric has been rearranged and a few additional fabric further. (iv) The bankruptcy on int~gral theorems, now bankruptcy five, has been multiplied to incorporate an altemative evidence of Gauss's theorem, a treatmeot of Green's theorem and a extra prolonged discussioo of the category of vector fields. (v) the one significant swap made in what are actually Chapters 6 and seven is the deletioo of the dialogue of the DOW out of date pot funetioo. (vi) A small a part of bankruptcy eight on Maxwell's equations has been rewritten to offer a fuller account of using scalar and veetor potentials in eleetromagnetic thought, and the devices hired were replaced to the m.k.s. system.

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**Example text**

X By these we now have that d = [e, d, b] a - [e, d, a] b ra, b, e] + ra, b, d] e , the denominator being non-zero since a, b, e are non-planar. Hence d = [b, e, d] a + [e, a, d] b + ra, b, d] e. ra, b, e] 7. LiBe and Surface IntegraJs as scalar Products. In fig. 18, let e be any eurve drawn in a veetor field and ds an element of are along it Ffgure 18 Tangentialline integrai of a vector at any point P. Let V denote the veetor at P in a direction making an angle 8 with that of the length element.

Thus, we may write eurlz V = (O~ _ O~x) k. By following a precisely similar method with the two remaining rectangles we find the other components of curl V to be curlll V Thus, = ( -OVX - -OVz) J• OZ ox II) = (OVz -oy - -OV OZ and curlx V • I. • (OVx curI V -_ (OVz - - OV -II ) 1+ - - OVo). OZ. 11 bis) Vx V" Vz Now take the vector product V x V of the operator V and the vector V, using rectangular co-ordinates. 18, we have or, Vx V = (i :x + j ~ + k ~) = (OVz _ OVII ) i oy that is, OZ x (V:ei + (OV:e + V,,) + Vzk) _ OVz) j OX V x V = curl V.

3. partial Difl'erentiatiOll. The simple properties of difl'erentiation as applied to veetors in §§ 1 and 2 can be extended to partial derivatives when a veetor is a function of more than one independent scalar variable. e. of a vector field. If y and z remain constant while x increases, the partial derivative oV/ox denotes the rate of increase of V with respeet to x. Likewise changing y and z alone we obtain the partial derivatives õV/oy and õV/oz, denoting the rates ofincrease with respeet to y and z, respeetively.

### An Introduction to Vector Analysis For Physicists and Engineers by B. Hague D.SC., PH.D., F.C.G.I. (auth.)

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