Double Integrals ?

Since the Electric field is constant, there is no charge enclosed by the solid paraboloid body. Hence by Gauss' law, the total flux computed over the entire body is zero, i.e., $\phi_{\text{curved surface}} + \phi_{\text{base}}=0$ Thus, $\phi_{\text{curved surface}} =- \phi_{\text{base}}$ The right hand side of the equation is readily computed to be $\phi_{\text{base}}=\pi R^2 E$ Thus, $|\phi_{\text{curved surface}} |=\pi R^2 E$

2 methods :

Easy method : $\Phi_{net} = 0$ ; $\Phi_{circular} = -\pi E_{0}R^2$ ( inward flux) hence we simply obtain $\Phi_{curved} = \pi E_{0}R^2$ .

Tough method : Focus on a differential area dA like in this figure

clearly

$dA = r^2 cos\theta d\theta d\phi$ ; For simplicity lets assume $\vec{E} = a \hat{z}$ .

The normal vector is resolved as

$\hat{r} = sin\theta \hat{z} + cos\theta sin \phi \hat{x} + cos\theta cos\phi \hat{y}$

Now using Gauss Law

$\Phi_{E}:=\oint \vec{E} \cdot \vec{r} dA$ clearly $\vec{E} \cdot \vec{r} = E_{0} sin \theta$

So now we need $\displaystyle\int \displaystyle\int E_{0} sin\theta R^2 cos \theta d\theta d\phi$

$\theta \in [0,\dfrac{\pi}{2}]$ and $\phi \in [0,\ 2\pi]$ so that

$\int_0^{\pi/2} \int_0^{2\pi} E_{0}R^2 sin\theta cos \theta d\theta d\phi = 2\pi E_{0}R^2 \int_0^{\pi/2} sin\theta cos \theta d\theta = \boxed{\pi E_{0}R^2}$

I'd look at sir @Abhishek Sinha's solution, 'cause it is so much simpler. But, if we want hard work, the flux $\Phi$ is defined by

$\Phi = \int \int_{R} \overrightarrow{F} \cdot \langle -f_x, -f_y, 1 \rangle \text{d}A,$

where $R$ is the region we're considering, $\overrightarrow{F}$ is our vector field and $f$ defines the surface. It is easy to notice that a possible choice for $\overrightarrow{F}$ and $f$ is

$\overrightarrow{F} = E \langle 0, 0, 1 \rangle; \;\;\;\; f(x,y) = \frac{x^2 + y^2}{4},$

where the region $R$ must be a circle of radius $2$ . Therefore

$\Phi = \int \int_{R} E \langle 0, 0, 1 \rangle \cdot \langle -x/2, -y/2, 1 \rangle \text{d}A = E \int \int_{R} \text{d}A = 4\pi E \approx 125.663 \text{Vm}.$