Electric field of an infinite sheet of charge I am trying to derive the formula for E due to an infinite sheet of charge with a charge density of $ \rho C/m^2$. I assumed the sheet is on $yz$-plane. I used Coulomb's law to get an equation and integrated the expression that over $yz$-plane. But, I have not succeeded in deriving the correct expression. The answer I am getting is $0$.
Below is the picture of my work. Kindly, have a look and let me know where did I make mistakes. 
In actual, E due to a charge sheet is constant and the correct expression is 
E $=\rho/2\epsilon$0 aN , where aN is unit vector normal to the sheet. 

 A: Method 1 (Gauss’ law):
Just simply use Gauss’ law:
$$\int_{\partial V}  \vec{E} \cdot \vec{da} = \frac{Q}{\epsilon_0}.$$ 
A pillbox using Griffiths’ language is useful to calculate $\vec{E}$. The pillbox has some area $A$. And due to symmetry we expect the electric field to be perpendicular to the infinite sheet. Imagine putting a test charge above it, in which way does it move? Right, perpendicular to the sheet. Using $Q=\rho A$ for the charge enclosed in the pillbox we get: 
$$ \rho A = \epsilon_0 \int_{\partial V} |\vec{E}| |\vec{da}| = \epsilon_0 \int_{\partial V} E da = \epsilon_0 E \int_{\partial V} da = \epsilon_0 (2AE), $$
since we expect $E$ to be constant for fixed distance for the infinite sheet. Note that the sides of the pillbox do not contribute to the integral since $\vec{E} \cdot \vec{da} = 0$ in that case. 
All together we find that $E=\frac{\rho}{2 \epsilon_0}$ and the direction we thought already of is some unit vector $\hat{n}$ orthogonal to the infinite sheet:
$$ \vec{E} = \frac{\rho}{2 \epsilon_0} \hat{n} .$$ 
Method 2: (Coulomb/direct calculation)
Another method goes as follows:
$$E=E_x= k \int \frac{x}{(r^2 + x^2)^{3/2}} r dr d\theta = 2\pi k \int \frac{xr}{(r^2 + x^2)^{3/2}} dr = 2\pi kx [ (r^2 + x^2)^{-1/2}]^0_{\infty} = 2\pi k x \frac{1}{x}=  2\pi k.$$ Let us see, I called $$k= \frac{\rho}{4 \epsilon_0 \pi}$$ we get indeed that $E=\frac{\rho}{2 \epsilon_0}$. 
Errors in your calculation: 
- the $y$ in the nominator should be a $x$.
- missing term in the denominator, namely $z^2$ because now you consider an infinite line and integrate over a surface. 
A: Are you looking to do the integrations by hand? Because 
$r = x \hat{x} + y \hat{y} + z \hat{z}$
$r^\prime = y^\prime \hat{y} + z^\prime \hat{z}$
Should yield the correct answer, but the integrations are messy, unless you go to cylindrical coordinates 
A: Use cylindrical coordinates. The field (on axis) of a ring of charge (radius $R$, charge density $\lambda$) goes like:
$$ E(z) = \frac{1}{2\epsilon_0}\frac{ R z}{(z^2 + R^2)^{\frac 3 2}} $$
then integrate over $R$, using:
$$ \int{\frac{ R z}{(z^2 + R^2)^{\frac 3 2}dR}}=-\frac z {\sqrt{z^2+R^2}}\rightarrow 1$$
