Story of Sinx and Flux

Consider a point on the X-Y plane having a position vector:

$\vec{r}_p = x \hat{i} + y\hat{j}$

The location of the charge is:

$\vec{r}_c = 3 \hat{i} + 3\hat{j}+ \hat{k}$

The vector joining these two points and directed away from the charge is:

$\vec{r} = \vec{r}_p - \vec{r}_c$

An elementary area around this point can be defined as:

$d\vec{S} = -dy \ dx \hat{k}$

The electric field at the point $P$ is:

$\vec{E} = \frac{q}{4 \pi} \frac{\vec{r}}{\lvert \vec{r} \rvert^3}$

The flux through the elementary area is:

$d\phi = \vec{E} \cdot d\vec{S}$

Substituting all expressions and simplifying and integrating leads to the following expression for flux through the given area:

$\boxed{\phi = \frac{q}{4 \pi} \left( \int_{0}^{\pi} \int_{0}^{\sin{x}} \frac{dy \ dx}{\left( \left(x-3\right)^2 + \left(y-2\right)^2 + 1\right)^{3/2}} + \int_{\pi}^{2\pi} \int_{\sin{x}}^{0} \frac{dy \ dx}{\left( \left(x-3\right)^2 + \left(y-2\right)^2 + 1\right)^{3/2}}\right) \approx 0.185}$