To find the volume of the solid under the surface $z = 42y^2 - 12x$ and above the region $R$ in the $xy$-plane, we set up a double integral over the region $R$.

Given the region $R$:

• $0 \leq x \leq 4$
• $(x-2)^2 \leq y \leq 6$

First, convert $(x-2)^2 \leq y$ to $y \geq (x-2)^2$.

The volume $V$ can be expressed as the double integral:

$V = \iint_R (42y^2 - 12x) \, dy \, dx$

Substitute the limits for integration:

$V = \int_{0}^{4} \int_{(x-2)^2}^{6} (42y^2 - 12x) \, dy \, dx$

Now, solve the inner integral with respect to $y$:

$\int_{(x-2)^2}^{6} (42y^2 - 12x) \, dy$

Calculate:

$= \int_{(x-2)^2}^{6} 42y^2 \, dy - \int_{(x-2)^2}^{6} 12x \, dy$

Solve each part separately:

1. $\int 42y^2 \, dy = 42 \left[ \frac{y^3}{3} \right] = 14y^3$
2. $\int 12x \, dy = 12xy$

Evaluating these from $y = (x-2)^2$ to $y = 6$:

For $14y^3$: