Question #192657

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


1
Expert's answer
2021-05-13T09:21:08-0400

divA=3x2i+yk,\text{div} \vec{A}=3x^2\vec{i}+y\vec{k},

Φ=02dx02dy02(3x2+y)dz=202dx02(3x2+y)dy=202(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.


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