Задачи по векторному анализу. Михайлов В.К - 103 стр.

îòêóäà
                               C1 = ρ cos ϕ + C2 (ϕ ) ,
ò. å.
                       u = ln ρ ⋅ arctg z + ρ cos ϕ + C2(ϕ ) .
Ïîäñòàâëÿÿ ïîñëåäíåå âûðàæåíèå âî âòîðîå óðàâíåíèå ñèñòåìû
(4.14), ïîëó÷àåì:
                                         ∂C2
                           − ρ sin ϕ +       = −ρ sin ϕ ,
                                         ∂ϕ
ò. å. Ñ2 = const = C. Òàêèì îáðàçîì, èñêîìûé ïîòåíöèàë
                     u( ρ, ϕ , z ) = ln ρ ⋅ arctg z + ρ cos ϕ + C .


                                     Çàäà÷è
     • Óñòàíîâèòü, ÿâëÿþòñÿ ëè ïîòåíöèàëüíûìè ñëåäóþùèå âåê-
òîðíûå ïîëÿ, è åñëè äà, òî âû÷èñëèòü èõ ïîòåíöèàëû ñ òî÷íîñ-
òüþ äî àääèòèâíîé ïîñòîÿííîé:
              ρ          ρ      ρ
        4.44. a =               eρ ;
                      ρ2 + 4
                ρ          ρ
        4.45.   a = f ( ρ )eρ ;
                ρ                   ρ            ρ  ρ
        4.46.   a = (2ρ + sin ϕ )eρ + cos ϕ eϕ + 2 zez ;
                ρ     ρ
        4.47.   a = ρeϕ ;
                ρ       ρ                   ρ
        4.48.   a = ρzeρ + ( ρ 2 / 2 − z 2 )ez ;
                ρ ρ
        4.49.   a = eϕ / ρ ;
              ρ            ρ eρ cos ϕ ρ       ρ
        4.50. a = eρ sin ϕ eρ +       eϕ + 2 zez ;
                                 ρ
              ρ    ρ ϕρ         ρ
        4.51. a = ρeρ + ρ eϕ + zez ;
              ρ         ρ
        4.52. a = f (r )er ;




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