I want to evaluate $$\iint_S\mathbf F\cdot d\mathbf S$$ where S is the top half of a unit sphere $S =\{(x,y,z): x^2+y^2+z^2\leq1, z\geq0\}$ and the vector field $$\mathbf F = \langle xz^2,\; \frac 1 3 y^3 + \arctan(z),\; x^2z+y^2 \rangle.$$
The way i figured i would approach this is i would seal the open dome with a unit disc $D =\{(x,y,0): x^2+y^2\leq1\}$ in the $xy$ plane so that it forms a simple solid $K$ region that i can calculate using the divergence theorem like this
$$\iint_S\mathbf F\cdot d\mathbf S + \iint_D\mathbf F\cdot d\mathbf S = \iiint_{K}\text{div}(\mathbf F)\, dV$$ $$\iint_S\mathbf F\cdot d\mathbf S = - \iint_D\mathbf F\cdot d\mathbf S + \iiint_{K}\text{div}(\mathbf F)\, dV$$
Parameterizing the volume using spherical coordinates i get
$$\text{div}(\mathbf F) = z^2+y^2+x^2 = r^2$$ $$\text{spherical correction term} = r^2 \sin \phi$$ $$ \iiint_{K}\text{div}(\mathbf F)\, dV= \int_0^{\pi/2} \int_0^{2\pi} \int_0^1 {r^2 \; r^2 \sin \phi \; dr \; d\theta \; d\phi} = \frac {2 \pi} 5 $$
and parameterizing the disc using polar coordinates i get $$\iint_D\mathbf F\cdot d\mathbf S = \int_0^{2\pi} \int_0^1 {\mathbf F(p(r, \theta)) \cdot normal \; dr \; d\theta}$$ $$p(r, \theta) = \langle r \cos \theta, r \sin \theta, 0 \rangle$$ $$normal = \langle 0, 0, r \rangle$$ $$\mathbf F(p(r, \theta)) \cdot normal = (r \sin \theta)^2 \; r $$ $$\int_0^{2\pi} \int_0^1 {(r \sin \theta)^2 \; r \; dr \; d\theta} = \frac \pi 4$$
So the sum of the disc and dome is $$ \frac {2 \pi} 5 - \frac \pi 4 = \frac {3 \pi} {20}$$
But the answer should be $\frac {13 \pi} {20}$ so where am i going wrong. I have double checked all the calculations of the integrals so im fairly certain that is not the problem. I have also tried other similar problems but i always get the wrong answer so i assume there is something im missing.