The value of surface integral using the Divergence Theorem is [tex]729\pi[/tex] .
Divergence Theorem states that the surface integral of a vector field over a closed surface, is equal to the volume integral of the divergence over the region inside the surface. Mathematically the it can be calculated using the formula:[tex]\int\int\int\limit{ }_V(\delta \cdot F)=\int\int(F \cdot n)dS[/tex]
The divergence of F is
[tex]div F=\frac{d}{dx}(2x^{3}+y^{3})+\frac{d}{dy}( y^{3} +z^{3})+\frac{d}{dz}3y^{3} z[/tex]
[tex]div F=6x^{2}+3y^{2}+3y^{2}[/tex]
Let E be the region [tex]{(x,y,z):0\leq z\leq 9-x^{2} -y^{2}[/tex] then by divergence theorem we have [tex]\int \int\limits^{}_s {F\cdot n\times dS} =\int\int\int\limits^{}_E divFdV=\int\int\int\limits^{}_E(6x^2+6y^2)dV[/tex]
Now we find the value of the integral:
[tex]=\int\limits^{2\pi}_0\int\limits^3_0\int\limits^{9-r^2}_0(6r^2)rdzdrd{\theta}\\=\int\limits^{2\pi}_0 \int\limits^3_0(9-r^2)6r^3drd{\theta}\\=2\pi\int\limits^3_0 {(54r^3-6r^5)} dr\\[/tex]
[tex]=2\pi\times \frac{729}{2}\\=729\pi[/tex]
Thus we can say that the value of the integral for the surface around the paraboloid is given by [tex]729\pi[/tex].
To learn more about Divergence Theorem :
https://brainly.com/question/17177764
