To find the surface area of the solid of revolution generated by revolving the curve $f(x) = -\cosh(1.65(x - 4.45)) + 9.3$ about the y-axis, we can use the following formula:

$\text{Surface Area} = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Step 1: Differentiate the function with respect to $x$

First, find the derivative of $f(x)$ with respect to $x$.

Given: $f(x) = -\cosh(1.65(x - 4.45)) + 9.3$

Differentiate $f(x)$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}\left[-\cosh(1.65(x - 4.45)) + 9.3\right]$

Using the chain rule: $\frac{dy}{dx} = -\sinh(1.65(x - 4.45)) \cdot 1.65$ $\frac{dy}{dx} = -1.65\sinh(1.65(x - 4.45))$

Step 2: Set up the integral

Substitute the expression for $\frac{dy}{dx}$ into the surface area formula:

$\text{Surface Area} = 2\pi \int_{a}^{b} x \sqrt{1 + \left(-1.65\sinh(1.65(x - 4.45))\right)^2} \, dx$

Simplifying the square: $\text{Surface Area} = 2\pi \int_{a}^{b} x \sqrt{1 + 1.65^2\sinh^2(1.65(x - 4.45))} \, dx$

Step 3: Define the limits of integration

To compute the surface area, you would typically need the limits of integration $[a, b]$, which represent the interval over which the function is being revolved. However, if no specific interval is given, the surface area can't be evaluated as a definite number. Suppose the interval is $[a, b] = [c, d]$.

Step 4: Evaluate the integral

Without the specific limits $a$ and $b$, we can't compute the exact surface area, but the process would involve evaluating the above integral over the interval $[a, b]$.

Final Surface Area Formula

Assuming you have limits $[a, b]$, the surface area can be numerically integrated using:

$\text{Surface Area} = 2\pi \int_{a}^{b} x \sqrt{1 + 2.7225\sinh^2(1.65(x - 4.45))} \, dx$

Explanation:

1. Function Differentiation: The derivative of $f(x)$ gives us the slope at any point, which is crucial for calculating the surface area.
2. Surface Area Formula: The formula $2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$ accounts for the revolving about the y-axis.
3. Simplification: The simplification inside the square root reduces the expression for easier calculation.

Would you like to calculate the surface area over a specific interval or need further explanation on any step?

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Related Questions:

1. What is the physical significance of the surface area of revolution?
2. How does the function $\cosh(x)$ behave, and why is it used in this context?
3. What are some applications of surface area calculations in real life?
4. How would the surface area change if the function were revolved around the x-axis instead?
5. Can you find the volume of the solid of revolution using a similar approach?

Tip:

For complex integrals, numerical methods like Simpson's rule or using software tools can help evaluate the integral accurately.