$$ Prove \int \int_S F•\mathbb{n} dS = 4πa^5~ where ~F = r²<x,y,z>, r² =x²+y²+z²$$ and $S$ is the boundary of the sphere with radius a.
I am not sure what error I am making in the calculation (I can get the answer from other method but I am not sure what error I am making here)
$$ 2\int \int r² <x,y,z> • \frac{<2x,2y,2z>}{2z} dxdy = 2 \int \int a² \frac{2a²}{2z} dxdy= 2\int \int a⁴/z dxdy= \int \int \frac{2a⁴}{z} dxdy ≠ 4πa^5$$
Can somebody point out the error?
Since the unit normal vector to surface is given by $\frac{\mathbf{x}}{|\mathbf{x}|}$ then it follows \begin{align} \iint_{a\mathbb{S}^2} \mathbf{F}\cdot \mathbb{n}\ dS(\mathbf{x}) =&\ \iint_{a\mathbb{S}^2} \frac{|\mathbf{x}|^2\mathbf{x}\cdot \mathbf{x}}{|\mathbf{x}|}\ dS(\mathbf{x})\\ =&\ a^3 \iint_{a\mathbb{S}^2} dS(\mathbf{x}) \\ =&\ a^3\operatorname{Area}(a\mathbb{S}^2) = a^3\cdot 4\pi a^2. \end{align}
Edit: To do it your way you need to first parametrize your surface (so you can express $z$ as a function of $x, y$). Moreover, let us split the integral into two pieces, the upper hemisphere $a\mathbb{S}_u$ and the lower hemisphere $a\mathbb{S}_l$. Let us use the graph parametrization of $\mathbb{S}_u$ given by $\Phi(x, y) = (x, y, \sqrt{a^2-x^2-y^2})$, then the normal vector $\mathbb{n}$ to the surface is given by \begin{align} \Phi_x\times \Phi_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 0 & \frac{-x}{\sqrt{a^2-x^2-y^2}}\\ 0 & 1 & \frac{-y}{\sqrt{a^2-x^2-y^2}} \end{vmatrix} = \frac{x\mathbf{i}+y\mathbf{j}}{\sqrt{a^2-x^2-y^2}}+\mathbf{k} \end{align} with $|\Phi_x\times \Phi_y| = \frac{a}{\sqrt{a^2-x^2-y^2}}$ which means \begin{align} \mathbb{n} = \frac{\Phi_x\times \Phi_y}{|\Phi_x\times \Phi_y|} = \frac{x\mathbf{i}}{a}+ \frac{y\mathbf{j}}{a}+ \frac{\sqrt{a^2-x^2-y^2}\mathbf{k}}{a} \end{align} and \begin{align} dS(\mathbf{x}) = \frac{a}{\sqrt{a^2-x^2-y^2}}dxdy \end{align}
Then it follows \begin{align} \iint_{a\mathbb{S}_u} \mathbf{F}\cdot \mathbb{n}\ dS(\mathbf{x}) =&\ \iint_{|x|^2+|y|^2\leq a^2} a^2(x, y, \sqrt{a^2-x^2-y^2}) \cdot (x, y, \sqrt{a^2-x^2-y^2})\ \frac{dxdy}{\sqrt{a^2-x^2-y^2}}\\ =&\ a^4 \iint_{|x|^2+|y|^2\leq a^2} \frac{dxdy}{\sqrt{a^2-x^2-y^2}} = a^4 \int^{2\pi}_0 \int^a_0 \frac{r}{\sqrt{a^2-r^2}}dr d\theta = 2\pi a^5. \end{align} Similarly, the integral over $a\mathbb{S}_l$ gives $2\pi a^5$.