To calculate the surface area of the region bounded by the graph of the hyperbolic cosine function, $y = \cosh(x)$, and the lines $x = -2$, $x = 2$, and $y = 0$, we will set up an integral for the area under $y = \cosh(x)$ over the interval $x \in [-2, 2]$.

Step 1: Setting Up the Integral

The surface area of the region is given by the integral of $y = \cosh(x)$ from $x = -2$ to $x = 2$, with respect to $x$.

$\text{Area} = \int_{-2}^{2} \cosh(x) \, dx$

Step 2: Integrate $\cosh(x)$

The hyperbolic cosine function $\cosh(x)$ has an antiderivative:

$\int \cosh(x) \, dx = \sinh(x)$

Thus, we can integrate $\cosh(x)$ from $-2$ to $2$:

$\text{Area} = \left[ \sinh(x) \right]_{-2}^{2} = \sinh(2) - \sinh(-2)$

Step 3: Simplify the Expression

Since $\sinh(-x) = -\sinh(x)$, we have:

$\text{Area} = \sinh(2) - (-\sinh(2)) = 2 \sinh(2)$

Step 4: Substitute $\sinh(2)$

Using the definition $\sinh(2) = \frac{e^2 - e^{-2}}{2}$:

$\text{Area} = 2 \cdot \frac{e^2 - e^{-2}}{2} = e^2 - e^{-2}$

Final Answer

The surface area of the region is:

$\text{Area} = e^2 - e^{-2}$

Let me know if you'd like further details or have any questions!

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1. How does the hyperbolic cosine differ from the regular cosine in terms of area under the curve?
2. Can this approach be applied to any region bounded by hyperbolic functions?
3. What would the surface area be if the bounds were extended to $x = -3$ and $x = 3$?
4. How do hyperbolic functions relate to exponential functions in area calculations?
5. What happens to the area if we change $y = 0$ to $y = c$ for a constant $c$?

Tip: The hyperbolic cosine function, unlike the regular cosine, does not oscillate and is always positive, making area calculations straightforward within specified bounds.