To find the surface area of the given trough, we use the formula for the surface area of a graph $z = f(x, y)$:

$S = \int \int_D \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dA$

Step 1: Partial Derivatives

The surface is given by $z = 4x^2$. First, calculate the partial derivatives:

• $\frac{\partial z}{\partial x} = 8x$
• $\frac{\partial z}{\partial y} = 0$

Step 2: Plug into the Surface Area Formula

The integrand becomes: $\sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} = \sqrt{1 + (8x)^2 + 0^2} = \sqrt{1 + 64x^2}.$

Thus, the surface area formula is: $S = \int \int_D \sqrt{1 + 64x^2} \, dx \, dy$ where the region $D$ is $-2 \leq x \leq 2$ and $0 \leq y \leq 2$.

Step 3: Separate the Integrals

Since the integrand depends only on $x$, we can separate the $x$- and $y$-integrals: $S = \int_0^2 \int_{-2}^2 \sqrt{1 + 64x^2} \, dx \, dy.$

The $y$-integral simply contributes a factor of 2: $\int_0^2 1 \, dy = 2.$

So, the surface area becomes: $S = 2 \int_{-2}^2 \sqrt{1 + 64x^2} \, dx.$

Step 4: Simplify the $x$-Integral

This integral cannot be solved in terms of elementary functions, but it can be evaluated as: $I = \int_{-2}^2 \sqrt{1 + 64x^2} \, dx.$

Using the symmetry of the integrand: $I = 2 \int_0^2 \sqrt{1 + 64x^2} \, dx.$

Step 5: Solve Using Substitution

Let $u = 1 + 64x^2$, so $du = 128x \, dx$. When $x = 0$, $u = 1$, and when $x = 2$, $u = 1 + 64(2^2) = 257$. The integral becomes: $\int_0^2 \sqrt{1 + 64x^2} \, dx = \frac{1}{64} \int_1^{257} \sqrt{u} \, du.$

Now evaluate: $\int_1^{257} \sqrt{u} \, du = \int_1^{257} u^{1/2} \, du = \left[\frac{2}{3} u^{3/2}\right]_1^{257}.$

$\int_1^{257} \sqrt{u} \, du = \frac{2}{3} \left[257^{3/2} - 1^{3/2}\right].$

Step 6: Final Expression for Surface Area

Substitute back into the surface area formula: $S = 2 \cdot \frac{1}{64} \cdot \frac{2}{3} \left[257^{3/2} - 1\right].$

Simplify: [ S = \frac{4}{192} \left[257^{3/2} - 1\right] = \frac{1}{48} \left[257^{3/2} - 1\right]. ]

Numerical Approximation

Using $257^{3/2} \approx 4122.0$: $S \approx \frac{1}{48} (4122.0 - 1) \approx \frac{1}{48} \cdot 4121.0 \approx 85.85.$

Thus, the surface area is approximately: $\boxed{85.85 \, \text{units}^2}.$

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Follow-up Questions

1. Would you like a step-by-step explanation of the substitution process?
2. Should I calculate the exact surface area using a numerical method?
3. Do you need help understanding why the symmetry simplifies the integral?
4. Would you like to see the geometric interpretation of the trough’s surface?
5. Should I explain alternative methods to compute this integral?

Tip: For problems involving surface areas of parametric surfaces, always start by finding the partial derivatives accurately.