By the divergence theorem, the flux of F across the boundary of a region, ∂R, is equal to the integral of div(F ) over the region itself, R. In this case, the flux would be
[tex]\displaystyle \iint_{\partial R}\mathbf F\cdot\,\mathrm d\mathbf S = \iiint_R\mathrm{div}(\mathbf F)\,\mathrm dx\,\mathrm dy\,\mathrm dz \\\\ = \int_0^1\int_0^3\int_0^1\frac{\partial(2x^2)}{\partial x} + \frac{\partial(5y^2)}{\partial y} + \frac{\partial(3z^2)}{\partial z} \,\mathrm dx\,\mathrm dy\,\mathrm dz \\\\ = \int_0^1\int_0^3\int_0^1 (4x+10y+6z)\,\mathrm dx\,\mathrm dy\,\mathrm dz = \boxed{60}[/tex]
In this exercise we have to calculate the flux by the divergent theorem:
[tex]\int\limits\int\limits_R\, ds = 60[/tex]
By the divergence theorem, the flux of F across the boundary of a region, ∂R, is equal to the integral of div(F ) over the region itself, R. In this case, the flux would be:
[tex]\int\limits\int\limits_R {F} \, ds= \int\limits \int\limits \int\limits_R {div(F)} \, dxdydz\\=\int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx dy dz= 60[/tex]
See more about vectorial calculus at : brainly.com/question/6960786
