Let f(x) = √x defined on the interval [4, 9].

1. Determine the inverse function f-1(y).

2. Find the surface area of the solid formed by revolving the region under the curve y = f(x) around the y-axis.

Calculate the integral to obtain the surface area of the solid of revolution.

Let's go step by step.

### 1. **Determine the inverse function \( f^{-1}(y) \) of \( f(x) = \sqrt{x} \).**

We are given that:
\[
f(x) = \sqrt{x}
\]
To find the inverse function \( f^{-1}(y) \), we need to express \( x \) in terms of \( y \).

Starting from \( y = f(x) \):
\[
y = \sqrt{x}
\]
Now, solve for \( x \):
\[
x = y^2
\]
Thus, the inverse function is:
\[
f^{-1}(y) = y^2
\]

Since the domain of \( f(x) \) is \( [4, 9] \), the range of \( f^{-1}(y) \) will be \( [2, 3] \) because \( \sqrt{4} = 2 \) and \( \sqrt{9} = 3 \). Therefore, the inverse function \( f^{-1}(y) = y^2 \) is defined for \( y \in [2, 3] \).

---

### 2. **Find the surface area of the solid formed by revolving the region under the curve around the y-axis.**

To calculate the surface area of a solid formed by revolving a curve \( x = g(y) \) around the \( y \)-axis, we use the following formula for surface area:

\[
A = 2\pi \int_{y_1}^{y_2} x(y) \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy
\]
In this case, the curve is \( x = f^{-1}(y) = y^2 \), and we are revolving around the \( y \)-axis between \( y = 2 \) and \( y = 3 \).

#### Step-by-step calculation:

1. **Find \( \frac{dx}{dy} \):**
\[
x = y^2 \quad \Rightarrow \quad \frac{dx}{dy} = 2y
\]

2. **Substitute into the surface area formula:**
\[
A = 2\pi \int_2^3 y^2 \sqrt{1 + (2y)^2} \, dy
\]
Simplifying the expression under the square root:
\[
A = 2\pi \int_2^3 y^2 \sqrt{1 + 4y^2} \, dy
\]

We now need to compute this integral.

---

#### Let's compute the integral step by step.

\[
A = 2\pi \int_2^3 y^2 \sqrt{1 + 4y^2} \, dy
\]

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Would you like me to explain the next steps for calculating the integral, or do you have any other questions?

Here are 5 related questions to expand your understanding:
1. How can you derive surface area formulas for solids of revolution around different axes?
2. What is the geometric meaning of a solid of revolution in calculus?
3. What is the significance of inverse functions in geometry and calculus?
4. How can you check the domain and range when finding inverse functions?
5. What other methods can be used to find surface areas for curves not involving a square root?

**Tip:** When solving integrals for surface areas, breaking down the problem step-by-step (finding derivatives, simplifying expressions) helps in managing complexity.

Let f(x) = √x defined on the interval [4, 9].
1. Determine the inverse function f-1(y).
2. Find the surface area of the solid formed by revolving the region under the curve y = f(x) around the y-axis. Calculate the integral to obtain the surface area of the solid of revolution.

Learn how to find the inverse function of f(x) = √x on the interval [4, 9] and calculate the surface area of the solid formed by revolving the curve y = f(x) around the y-axis. This problem involves key calculus concepts, including inverse functions and surface areas of solids of revolution. Suitable for advanced high school students in Grades 11-12.

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