To find the surface area of the given surface $z = 4x^2$ for the bounds $-2 \leq x \leq 2$ and $0 \leq y \leq 6$, we use the surface area formula for a function $z = f(x, y)$ over a region $R$ in the $xy$-plane:

$\text{Surface Area} = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA$

Since $z = 4x^2$, let's find the partial derivatives with respect to $x$ and $y$:

1. Partial derivative with respect to $x$: $\frac{\partial z}{\partial x} = \frac{d}{dx}(4x^2) = 8x$

2. Partial derivative with respect to $y$: $\frac{\partial z}{\partial y} = 0$ (since $z$ does not depend on $y$).

Now, substitute these into the surface area formula:

$\text{Surface Area} = \iint_R \sqrt{1 + (8x)^2 + 0^2} \, dA = \iint_R \sqrt{1 + 64x^2} \, dx \, dy$

Setting up the Integration

The region $R$ in the $xy$-plane is defined by $-2 \leq x \leq 2$ and $0 \leq y \leq 6$. Thus,

$\text{Surface Area} = \int_0^6 \int_{-2}^2 \sqrt{1 + 64x^2} \, dx \, dy$

We can separate the integrals because the integrand does not depend on $y$:

$\text{Surface Area} = \int_0^6 \, dy \int_{-2}^2 \sqrt{1 + 64x^2} \, dx$

Evaluating the $y$-Integral

Since the integrand is independent of $y$, the $y$-integral contributes a factor of 6:

$\text{Surface Area} = 6 \int_{-2}^2 \sqrt{1 + 64x^2} \, dx$

Solving the $x$-Integral

To evaluate $\int_{-2}^2 \sqrt{1 + 64x^2} \, dx$, we can use symmetry since the integrand is even:

$\int_{-2}^2 \sqrt{1 + 64x^2} \, dx = 2 \int_0^2 \sqrt{1 + 64x^2} \, dx$

To solve this integral, we can use a trigonometric substitution or consult a table of integrals. Using the substitution $x = \frac{1}{8} \tan(\theta)$, the integral evaluates to:

$2 \int_0^2 \sqrt{1 + 64x^2} \, dx = \left[ \frac{1}{24} \sinh^{-1}(8x) \right]_0^2$

Finally, the exact area can be computed