Close off the hemisphere
by attaching to it the disk
of radius 3 centered at the origin in the plane
. By the divergence theorem, we have
![\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS%5Ccup%20D%7D%5Cvec%20F%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cvec%20S%3D%5Ciiint_R%5Cmathrm%7Bdiv%7D%5Cvec%20F%28x%2Cy%2Cz%29%5C%2C%5Cmathrm%20dV)
where
is the interior of the joined surfaces
.
Compute the divergence of
:
![\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2](https://tex.z-dn.net/?f=%5Cmathrm%7Bdiv%7D%5Cvec%20F%28x%2Cy%2Cz%29%3D%5Cdfrac%7B%5Cpartial%28xz%5E2%29%7D%7B%5Cpartial%20x%7D%2B%5Cdfrac%7B%5Cpartial%5Cleft%28%5Cfrac%7By%5E3%7D3%2B%5Csin%20z%5Cright%29%7D%7B%5Cpartial%20y%7D%2B%5Cdfrac%7B%5Cpartial%28x%5E2z%2By%5E2%29%7D%7B%5Cpartial%20k%7D%3Dz%5E2%2By%5E2%2Bx%5E2)
Compute the integral of the divergence over
. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:
![\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%28%5Crho%2C%5Ctheta%2C%5Cvarphi%29%3D%5Crho%5Ccos%5Ctheta%5Csin%5Cvarphi%5C%5Cy%28%5Crho%2C%5Ctheta%2C%5Cvarphi%29%3D%5Crho%5Csin%5Ctheta%5Csin%5Cvarphi%5C%5Cz%28%5Crho%2C%5Ctheta%2C%5Cvarphi%29%3D%5Crho%5Ccos%5Cvarphi%5Cend%7Bcases%7D%5Cimplies%5Cbegin%7Bcases%7Dx%5E2%2By%5E2%2Bz%5E2%3D%5Crho%5E2%5C%5C%5Cmathrm%20dV%3D%5Crho%5E2%5Csin%5Cvarphi%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi%5Cend%7Bcases%7D)
So the volume integral is
![\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_Rx%5E2%2By%5E2%2Bz%5E2%5C%2C%5Cmathrm%20dV%3D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E3%5Crho%5E4%5Csin%5Cvarphi%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi%3D%5Cfrac%7B486%5Cpi%7D5)
From this we need to subtract the contribution of
![\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_D%5Cvec%20F%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cvec%20S)
that is, the integral of
over the disk, oriented downward. Since
in
, we have
![\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20F%28x%2Cy%2C0%29%3D%5Cdfrac%7By%5E3%7D3%5C%2C%5Cvec%5Cjmath%2By%5E2%5C%2C%5Cvec%20k)
Parameterize
by
![\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%28u%2Cv%29%3Du%5Ccos%20v%5C%2C%5Cvec%5Cimath%2Bu%5Csin%20v%5C%2C%5Cvec%5Cjmath)
where
and
. Take the normal vector to be
![\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%5Cvec%20r%7D%7B%5Cpartial%20v%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cvec%20r%7D%7B%5Cpartial%20u%7D%3D-u%5C%2C%5Cvec%20k)
Then taking the dot product of
with the normal vector gives
![\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v](https://tex.z-dn.net/?f=%5Cvec%20F%28x%28u%2Cv%29%2Cy%28u%2Cv%29%2C0%29%5Ccdot%28-u%5C%2C%5Cvec%20k%29%3D-y%28u%2Cv%29%5E2u%3D-u%5E3%5Csin%5E2v)
So the contribution of integrating
over
is
![\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E3-u%5E3%5Csin%5E2v%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D-%5Cfrac%7B81%5Cpi%7D4)
and the value of the integral we want is
(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>
==> 486π/5 - (-81π/4) = 2349π/20