Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of the sphere x2 + y2 + z2 = 1. (Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x2 + y2 ≤ 1, oriented downward, and S2 = S1 ∪ S.)

Looks like we have[tex]\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k[/tex]which has divergence[tex]\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2[/tex]By the divergence theorem, the integral of [tex]\vec F[/tex] across [tex]S[/tex] is equal to the integral of [tex]\nabla\cdot\vec F[/tex] over [tex]R[/tex], where [tex]R[/tex] is the region enclosed by [tex]S[/tex]. Of course, [tex]S[/tex] is not a closed surface, but we can make it so by closing off the hemisphere [tex]S[/tex] by attaching it to the disk [tex]x^2+y^2\le1[/tex] (call it [tex]D[/tex]) so that [tex]R[/tex] has boundary [tex]S\cup D[/tex].Then by the divergence theorem,[tex]\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV[/tex]Compute the integral in spherical coordinates, setting[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]so that the integral is[tex]\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5[/tex]The integral of [tex]\vec F[/tex] across [tex]S\cup D[/tex] is equal to the integral of [tex]\vec F[/tex] across [tex]S[/tex] plus the integral across [tex]D[/tex] (without outward orientation, so that[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S[/tex]Parameterize [tex]D[/tex] by[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath[/tex]with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex]. Take the normal vector to [tex]D[/tex] to be[tex]\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k[/tex]Then we have[tex]\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4[/tex]Finally,[tex]\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}[/tex]