I attempted to calculate the surface integral $\iint_S F \cdot n \ dS$ where $F = (x^2 + y^2 + z^2)(x, y, z)$ and $S$ is the sphere of radius $a$ centered at the origin. I used the divergence theorem to obtain the result, but it doesn't coincide with the result I obtain via the definition of the surface integral and I cannot find my mistake.
Clearly $\nabla \cdot F = 5(x^2 + y^2 + z^2)$, so by doing a change to spherical coordinates we get \begin{align*} \iint_S F \cdot n \ dS &= \iiint_{x^2 + y^2 + z^2 = a^2} 5(x^2 + y^2 + z^2) \ dV \\ &= \int_0^{2\pi} \int_0^\pi \int_0^a 5a^2\rho^2 \sin \phi \ d\rho d\phi d\theta \\ &= 5a^2 \int_0^{2\pi} \int_0^\pi \frac{a^3}{3} \sin \phi \ d\phi d\theta \\ &= \frac{10}{3} \pi a^5 \int_0^\pi \sin \phi \ d\phi \\ &= \frac{20}{3} \pi a^5. \end{align*}
Now, let's use the definition directly to compute the surface integral.
First, let's parametrize the sphere via spherical coordinates:
$$x = a \sin \phi \cos \theta, \quad y = a \sin \phi \sin \theta, \quad z = a \cos \phi.$$
So, our vector field, in this parametrization is equal to:
$$F = a^3(\sin \phi \cos \theta, sin \phi \sin \theta, \cos \phi).$$
And an outward pointing normal vector field is given by:
$$n = a^2 (\sin^2 \phi \cos \theta, \sin^2 \phi \sin \theta, \cos \phi \sin \phi).$$
By definition, the integrand of the double integral is the dot product of the previous two quantities:
$$F \cdot n = a^5 (\sin^3 \phi \cos^2 \theta + \sin^3 \phi \sin^2 \theta + \cos^2 \phi \sin \phi) = a^5 \sin \phi.$$
So, our surface integral is:
$$\iint_S F \cdot n \ dS = \int_0^\pi \int_0^{2\pi} a^5 \sin \phi \ d\theta d\phi = 2\pi a^5 \int_0^\pi \sin \phi \ d \phi = 4\pi a^5.$$
The results are not the same, and I cannot find my mistake.
Thanks in advance.