Answer to Question #157645 in Physics for Priya

Question

Using Gauss's theorem, calculate the flux of the vector field A vector=x³i^+x²zj^+yzk^ through the surface of a cube of side 2 units.

Answer 1

$$div {\vec{A}}=3x^2+0+y=3x^2+y\\ \Phi=\int_0^2\int_0^2\int_0^2(3x^2+y)dxdydz\\ \Phi=2\int_0^2\int_0^2(3x^2+y)dxdy\\ \Phi=2\int_0^2((2)3x^2+0. 5(2)^2)dx=40\\$$