Using guass theorem calculate the flux of the vector field A =x³i+x²zj+yzk through the square of a cube of a side 2 unit's
divA⃗=3x2i⃗+yk⃗,\text{div} \vec{A}=3x^2\vec{i}+y\vec{k},divA=3x2i+yk,
Φ=∫02dx∫02dy∫02(3x2+y)dz=2∫02dx∫02(3x2+y)dy=2∫02(6x2+2)dx=2(2x3+2x)∣02=40 Wb.\Phi=\int _0^2dx\int_0^2dy\int _0^2(3x^2+y)dz=2\int_0^2dx\int_0^2(3x^2+y)dy=2\int_0^2(6x^2+2)dx=2(2x^3+2x)|_0^2=40~Wb.Φ=∫02dx∫02dy∫02(3x2+y)dz=2∫02dx∫02(3x2+y)dy=2∫02(6x2+2)dx=2(2x3+2x)∣02=40 Wb.
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