divergencia y rotacional
TRANSCRIPT
CALCULO VECTORIAL
EN EL PLANO
πΉ(π₯, y)= 2xy π + sen(4xy2) π
EN EL ESPACIO
πΉ(π₯, y, z)=
π2π₯βln π₯π§ π + ctg(y2 β 9π§π₯β8) π β 3 cos π§ π
π
ππ₯π π₯, y = ππ₯ π₯, y π
π
ππ₯π π₯, y, π§ = ππ₯ π₯, y, π§ β π
π» =π
ππ₯ π +
π
ππ¦ π
π» =π
ππ₯ π +
π
ππ¦ π +
π
ππ§π
EN EL PLANO
πΉ π₯, y = π π + π π
div πΉ π₯, y = π» β πΉ π₯, y
EN EL ESPACIO
πΉ π₯, y = π₯ cos π¦ π + 2π₯π¦ π
SOLUCION
div πΉ π₯, y = π» β πΉ π₯, y
=π
ππ₯ π +
π
ππ¦ π β π₯ cos π¦ π + 2π₯π¦ π
=π
ππ₯π₯ cos π¦ +
π
ππ¦2π₯π¦
div πΉ π₯, y = cos π¦ + 2π₯
πΉ π₯, y,z = π₯ β π ππ π¦ π + 2 cos π¦ π + π₯ ln π§ π
SOLUCION:
div πΉ π₯, y, π§ = π» β πΉ π₯, y,z
=π
ππ₯ π +
π
ππ¦ π +
π
ππ§π β π₯ β π ππ π¦ π + 2 cos π¦ π + π₯ ln π§ π
=π
ππ₯π₯ β π ππ π¦ +
π
ππ¦2 cos π¦ π +
π
ππ§π₯ ln π§
div πΉ π₯, y, π§ = 1 β 2 senπ¦ +π₯
π§
πππ πΉ π₯, y,z = π» Γ πΉ π₯, y,z
=
π π ππ
ππ₯
π
ππ¦
π
ππ§π π π
=π
ππ¦
π
ππ§π π
π βπ
ππ₯
π
ππ§π π
π +π
ππ₯
π
ππ¦π π
π§
πΉ π₯, y, z = 34π₯y β π πππ₯ π + π₯π§ π + (4yβ8x)π
SOLUCION:
πππ πΉ π₯, y,z = π» Γ πΉ π₯, y,z =
π π ππ
ππ₯
π
ππ¦
π
ππ§
34π₯y β π ππ π₯ π₯π§ 4y β 8π₯
=
π
ππ¦
π
ππ§
π₯π§ 4y β 8π₯
π β
π
ππ₯
π
ππ§34π₯y β π ππ π₯ 4y β 8π₯
π +
π
ππ₯
π
ππ¦
34π₯y β π ππ π₯ π₯π§
π§
=π
ππ¦4π¦ β 8π₯ β
π
ππ§π₯π§ π β
π
ππ₯4y β 8π₯ β
π
ππ§34π₯y β π ππ π₯ π
+π
ππ₯π₯π§ β
π
ππ¦34π₯y β π ππ π₯ π
πππ πΉ π₯, y,z = 4 β π₯ π β β8π₯ π + π§ β 34π₯ π = 4 β π₯ π + 8π₯ π + π§ β 34π₯ π
πΉ π₯, y, z = 16π₯yπ§ π + y2 β π₯2 π + (3xz)π
SOLUCION:
πππ πΉ π₯, y,z = π» Γ πΉ π₯, y,z =
π π ππ
ππ₯
π
ππ¦
π
ππ§
16π₯yπ§ y2 β π₯2 3π₯π§
=
π
ππ¦
π
ππ§
y2 β π₯2 3π₯π§
π β
π
ππ₯
π
ππ§16π₯yπ§ 3π₯π§
π +
π
ππ₯
π
ππ¦
16π₯yπ§ y2 β π₯2
π§
=π
ππ¦3π₯π§ β
π
ππ§y2 β π₯2 π β
π
ππ₯3π₯π§ β
π
ππ§16π₯yz π +
π
ππ₯y2 β π₯2 β
π
ππ¦16π₯yz π
πππ πΉ π₯, y,z = 0 π β 3π§ β 16π₯y π + β2π₯ β 16π₯π§ π
= 0 π + 16π₯y β 3π§ π + β2π₯ β 16π₯π§ π