Question

c) Using Gauss' theorem, calculate the flux of the vector field Ā= x3 + x2zj + yzk
through the surface of a cube of side 2 units.
(10)​

Answer 1

Answer:

div{\vec{A}}=3x^2+y\\\Phi=\int_0^2\int_0^2\int_0^2(3x^2+y)dxdydz \\ div{\vec{A}}=3x^2+y\\\Phi=2^2\int_0^2(3x^2)dx+2^2\int_0^2(y)dy\\\Phi=2^2(2^3)+2^2(0.5\cdot2^2)\\\Phi=40divA=3x2+yΦ=∫02∫02∫02(3x2+y)dxdydzdivA=3x2+yΦ=22∫02(3x2)dx+22∫02(y)dyΦ=22(23)+22(0.5⋅22)Φ=40