Denote the unit sphere by [tex]\mathcal S[/tex]. If the sphere is closed, then the flux of [tex]\mathbf v(x,y,z)[/tex] across [tex]\mathcal S[/tex] is given by the divergence theorem to be
[tex]\displaystyle\iint_{\mathcal S}\mathbf v\cdot\mathrm d\mathbf S=\iiint_{\mathcal B}\nabla\cdot\mathbf v\,\mathrm dV[/tex]
where [tex]\mathcal B[/tex] denotes the space with boundary [tex]\mathcal S[/tex]. We have
[tex]\nabla\cdot\mathbf v=\dfrac{\partial(7xy^2)}{\partial x}+\dfrac{\partial(4x^2y)}{\partial y}+\dfrac{\partial(z^3)}{\partial z}=7y^2+4x^2+3z^2[/tex]
So the flux across [tex]\mathcal S[/tex] is equivalent to
[tex]\displaystyle\iiint_{\mathcal B}(7y^2+4x^2+3z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
We convert to spherical coordinates to evaluate the integral:
[tex]x=\rho\cos\theta\sin\varphi[/tex]
[tex]y=\rho\sin\theta\sin\varphi[/tex]
[tex]z=\rho\cos\varphi[/tex]
[tex]\implies 4x^2+7y^2+3z^2=3\rho^2+x^2+4y^2[/tex]
[tex]=3\rho^2+\rho^2\cos^2\theta\sin^2\varphi+4\rho^2\sin^2\theta\sin^2\varphi[/tex]
[tex]=\rho^2(3+\sin^2\varphi(1+3\sin^2\theta))[/tex]
[tex]\displaystyle\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^2(3+\sin^2\varphi(1+3\sin^2\theta))\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
[tex]=\dfrac{56\pi}{15}[/tex]
