Why does my polar integral contradict Gauss's law for a finite-sized charged disk?

Along disk carries a charge density that is proportional to the distance from the axis: $\rho = k r$, for some constant k. Find the electric field inside and outside this disk (radius is R).

Assume observation point is on the y-axis \begin{pmatrix} 0\\ a \end{pmatrix}

And random point from disk has \begin{pmatrix} r \cos \theta\\ r \sin \theta \end{pmatrix}

So the vector is \begin{pmatrix} -r \cos \theta\\ a-r \sin \theta \end{pmatrix}

$$E_x = \frac{1} {4 \pi \epsilon_0} \int_0^{2\pi} \!\!\!\int_0^R \frac{k\,r^{2}\,\bigl(-r \cos\theta\bigr)} {\Bigl[r^2 \cos^2 \theta + (\,a - r \sin\theta\,)^2\Bigr]^{3/2}} \,\mathrm{d}r\,\mathrm{d}\theta = 0 $$

In Polar coordinates,

$$ \frac{1} {4 \pi \epsilon_0} \int_0^{2\pi} \!\!\!\int_0^R \frac{k\,r^{2}\,\bigl(a - r \sin\theta\bigr)} {\Bigl[r^2 \cos^2 \theta + (\,a - r \sin\theta\,)^2\Bigr]^{3/2}} \,\mathrm{d}r\,\mathrm{d}\theta, $$

In Cartesian coordinates,

$$ \frac{1} {4 \pi \epsilon_0} \int_{-R}^R \int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}} \frac{k \sqrt{x^2 + y^2} (a - y)} {(x^2 + (a - y)^2)^{3/2}}\, dy \, dx $$

Case 1 when $0
Case 2 when $0
I am unable to solve this integral since it's divergent.

However, it can be solved symbolically if you use Gauss's law.

A much simpler scenario (from the standpoint of finding $\mathbf{E}$) is if the disk is actually a solid cylinder of radius $R$ and (say) thickness $t$. Suppose its volume‐charge density in cylindrical coordinates $(\rho,\phi,z)$ is $$ \rho_{\mathrm{vol}}(\rho) \;=\;k\,\rho. $$ Then by symmetry, the electric field inside must be purely radial (no dependence on $\phi$), and one can apply Gauss’s law in 3D very straightforwardly:

Hence Gauss’s law says $$ E(a)\,\bigl(2\pi\,a\,t\bigr) \;=\;\frac{Q_{\text{enclosed}}}{\epsilon_{0}}, $$ where the enclosed charge is $$ Q_{\text{enc}} \;=\;\int_{0}^{a}\!\!\int_{0}^{2\pi}\!\!\int_{0}^{t} k\,\rho\;\bigl(\rho\,\mathrm{d}\rho\,\mathrm{d}\phi\,\mathrm{d}z\bigr) \;=\;k\int_{0}^{a}\!\!\rho^{2}\,\mathrm{d}\rho \int_{0}^{2\pi}\!\mathrm{d}\phi \int_{0}^{t}\!\mathrm{d}z \;=\;k\,(2\pi)\,t\;\int_{0}^{a}\!\!\rho^{2}\,\mathrm{d}\rho \;=\;2\pi k\,t\;\frac{a^{3}}{3}. $$ Thus $$ E(a) \;=\;\frac{Q_{\text{enc}}}{2\pi\,a\,t\,\epsilon_{0}} \;=\;\frac{\tfrac{2\pi\,k\,t}{3}\,a^{3}}{2\pi\,a\,t\,\epsilon_{0}} \;=\;\frac{k\,a^{2}}{3\,\epsilon_{0}}, \quad (\,0\;\le a\le R\,). $$ That is a nice, simple, algebraic formula: the field inside grows like $a^{2}$. Outside the cylinder $(a>R)$, you do the same Gauss’s‐law reasoning but enclose the entire charge (radius $R$), and you find $$ E(a)\;=\;\frac{k\,R^{3}}{3\,\epsilon_{0}\,a} \quad\text{for }a \ge R. $$ All of this is much simpler precisely because in 3D you can use (true) Gauss’s law with a volume‐charge distribution that depends only on $\rho$.

But this answer from Gauss's law contradicts my polar integral answer when I am using numeric solutions.

Are my integral correct in both coordinates?

Your error is in the assumptions:

Both of these assumptions are false, since $\mathbf{E}$ is not in general horizontal.

Let's assume that the disk spans the range $z = [-t/2, t/2]$. For any $z$-coordinate other than $z = 0$, the electric field will have a $z$-component pointing away from the $xy$-plane. For example, consider the point $(0,0,t/2)$. It is not hard to see that the net electric field at this point will point along the axis (by symmetry), and that every charge element in your finite disk will give a positive contribution to $E_z$. So the end cap does contribute to the flux integral and cannot be disregarded.

More generally, for any point $P$ with a $z$-coordinate $\zeta$ between 0 and $t/2$, we can see that the contributions to $E_z$ from the charges between the planes $z = 2\zeta - t/2$ and $z = t/2$ will cancel out by symmetry (the red shaded region below cancels out with the blue shaded region.) But the charges outside of this region (the green shaded region) will provide a non-zero upwards contribution, since they all have $z$-coordinates less than $\zeta$. So there will be an upwards component to $E_z$ at all points with $z > 0$.

Similarly, there will be a downwards component to $E_z$ at all points with $z < 0$. In particular, there are non-zero $z$-components to the electric field at all points on the cylindrical side surface of your Gaussian surface, except for those points with $z = 0$; and so your first assumption about the direction of $\mathbf{E}$ is incorrect as well.

Note that in the more "common" case of an infinitely long cylinder (see, for example, Example 2.4 of Griffiths), $\mathbf{E}$ is purely radial. In that case, we can argue from symmetry that the electric field at all points is purely radial, since for any plane perpendicular to the cylinder axis there is an equal amount of charge on either side, and so by symmetry the electric field must lie in that plane. This is not the case for a finite cylinder like the one you're considering here, and so you cannot easily use Gauss's Law.