To find the surface area of the trough defined by $z = 3x^2$, over the region $-2 \leq x \leq 2$ and $0 \leq y \leq 4$, we use the formula for the surface area of a surface $z = f(x, y)$:

$S = \int \int_{R} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy$

Step-by-step solution:

1. Compute partial derivatives:

   • $\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(3x^2) = 6x$
   • $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(3x^2) = 0$

2. Substitute into the formula:

   • $1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = 1 + (6x)^2 + 0^2 = 1 + 36x^2$

3. Set up the surface area integral: $S = \int_{0}^{4} \int_{-2}^{2} \sqrt{1 + 36x^2} \, dx \, dy$

4. Simplify: The function $\sqrt{1 + 36x^2}$ depends only on $x$, so we can separate the integral: $S = \int_{0}^{4} \, dy \int_{-2}^{2} \sqrt{1 + 36x^2} \, dx$ The outer integral over $y$ is simply the length of $y$’s interval, which is $4$: $S = 4 \int_{-2}^{2} \sqrt{1 + 36x^2} \, dx$

5. Evaluate the integral: Let $u = 1 + 36x^2$, so $du = 72x \, dx$. Adjusting for substitution:

   • When $x = -2$, $u = 1 + 36(-2)^2 = 145$
   • When $x = 2$, $u = 1 + 36(2)^2 = 145$ The integral becomes: $\int_{-2}^{2} \sqrt{1 + 36x^2} \, dx = 2 \int_{1}^{145} \frac{\sqrt{u}}{72} \, du$ Simplify further: $= \frac{2}{72} \int_{1}^{145} u^{1/2} \, du = \frac{1}{36} \cdot \left[ \frac{2}{3} u^{3/2} \right]_{1}^{145}$ Evaluate: $= \frac{1}{36} \cdot \frac{2}{3} \left[ (145)^{3/2} - (1)^{3/2} \right]$ [ = \frac{2}{108} \left[ (145)^{3/2} - 1 \right] = \frac{1}{54} \left[ (145)^{3/2} - 1 \right] ]

6. Multiply by 4: [ S = 4 \cdot \frac{1}{54} \left[ (145)^{3/2} - 1 \right] = \frac{4}{54} \left[ (145)^{3/2} - 1 \right] ] Simplify: $S = \frac{2}{27} \left[ (145)^{3/2} - 1 \right]$

Final Answer:

$\boxed{\frac{2}{27} \left[ (145)^{3/2} - 1 \right]}$

Let me know if you would like further clarification or a numerical approximation.

Here are related questions:

1. Can we calculate this surface area numerically?
2. What if the trough extended to a different range for $x$?
3. How does the calculation change if $z$ were dependent on $y$ as well?
4. Can we interpret this surface area geometrically?
5. How is this integral simplified using symmetry?

Tip: Always verify if symmetry can reduce computation in double integrals for functions like this!