The graph of equation is a line
The surface is closed, so you can use the divergence theorem:
![\vec f=x^3\,\vec\imath+y^3\,\vec\jmath+z^3\,\vec k\implies\nabla\cdot\vec f=3x^2+3y^2+3z^2](https://tex.z-dn.net/?f=%5Cvec%20f%3Dx%5E3%5C%2C%5Cvec%5Cimath%2By%5E3%5C%2C%5Cvec%5Cjmath%2Bz%5E3%5C%2C%5Cvec%20k%5Cimplies%5Cnabla%5Ccdot%5Cvec%20f%3D3x%5E2%2B3y%5E2%2B3z%5E2)
Then the flux is
![\displaystyle\iint_S\vec f\cdot\mathrm d\vec S=3\iiint_Rx^2+y^2+z^2\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cvec%20f%5Ccdot%5Cmathrm%20d%5Cvec%20S%3D3%5Ciiint_Rx%5E2%2By%5E2%2Bz%5E2%5C%2C%5Cmathrm%20dV)
where
is the region with boundary
.
Convert to cylindrical coordinates, taking
![x=u\cos v](https://tex.z-dn.net/?f=x%3Du%5Ccos%20v)
![y=u\sin v](https://tex.z-dn.net/?f=y%3Du%5Csin%20v)
![z=z](https://tex.z-dn.net/?f=z%3Dz)
Then the integral is
![\displaystyle3\int_0^{2\pi}\int_0^1\int_{u^2}^5(u^2+z^2)u\,\mathrm dz\,\mathrm du\,\mathrm dv=\frac{525\pi}4](https://tex.z-dn.net/?f=%5Cdisplaystyle3%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E1%5Cint_%7Bu%5E2%7D%5E5%28u%5E2%2Bz%5E2%29u%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cfrac%7B525%5Cpi%7D4)
The answer is 480 , I need the points a lot please !!!