Define Q as the region bounded by the functions f(x)=x ^2/3 and g(x)=2x in the first quadrant between y=2 and y=3. If Q is rotated around the y-axis, what is the volume of the resulting solid?

Submit an exact answer in terms of π.

To calculate the volume of the solid formed by rotating the region \( Q \) around the \( y \)-axis, we use the method of cylindrical shells.

### Step 1: Identify the region and functions
The functions are:

- \( f(x) = x^{2/3} \)
- \( g(x) = 2x \)

The region \( Q \) is bounded in the first quadrant between \( y = 2 \) and \( y = 3 \), so the limits for \( y \) are \( y = 2 \) and \( y = 3 \).

### Step 2: Express \( x \) as a function of \( y \)

For \( f(x) = x^{2/3} \):
\[
y = x^{2/3} \quad \Rightarrow \quad x = y^{3/2}
\]

For \( g(x) = 2x \):
\[
y = 2x \quad \Rightarrow \quad x = \frac{y}{2}
\]

### Step 3: Volume using cylindrical shells formula
The volume of the solid formed by rotating the region around the \( y \)-axis is given by the formula:
\[
V = 2\pi \int_{y_1}^{y_2} \left( \text{radius} \times \text{height} \right) \, dy
\]
Here, the radius is \( x \), and the height is the difference between the functions \( g(x) \) and \( f(x) \).

The radius is the distance from the \( y \)-axis, which is simply \( x \), and the height is \( g(x) - f(x) \).

Substitute the expressions for \( x \) in terms of \( y \):
- For \( g(x) = 2x \), \( x = \frac{y}{2} \)
- For \( f(x) = x^{2/3} \), \( x = y^{3/2} \)

Thus, the volume integral becomes:
\[
V = 2\pi \int_2^3 \left( \frac{y}{2} - y^{3/2} \right) y \, dy
\]

### Step 4: Simplify and compute the integral
Simplify the integrand:
\[
V = 2\pi \int_2^3 \left( \frac{y^2}{2} - y^{5/2} \right) \, dy
\]

Now compute the integral term by term:
\[
V = 2\pi \left[ \frac{y^3}{6} - \frac{2}{7} y^{7/2} \right]_2^3
\]

### Step 5: Evaluate the definite integral
Evaluate at the upper and lower bounds \( y = 3 \) and \( y = 2 \):

For \( y = 3 \):
\[
\frac{3^3}{6} = \frac{27}{6} = 4.5, \quad \frac{2}{7} \times 3^{7/2} = \frac{2}{7} \times 27\sqrt{3}
\]

For \( y = 2 \):
\[
\frac{2^3}{6} = \frac{8}{6} = \frac{4}{3}, \quad \frac{2}{7} \times 2^{7/2} = \frac{2}{7} \times 8\sqrt{2}
\]

Now subtract the lower bound result from the upper bound result and compute the exact value in terms of \( \pi \).

Would you like further details on these calculations or clarifications? 

Here are 5 related questions:
1. How does the cylindrical shell method differ from the disk method for volume calculation?
2. Why is the integral set up with respect to \( y \) instead of \( x \)?
3. How would the method change if the region were rotated around the \( x \)-axis?
4. What are the practical uses of calculating volumes of solids of revolution?
5. How can the cylindrical shells method be applied to more complex curves?

**Tip**: Always sketch the region to better visualize the problem before starting volume calculations.

Define Q as the region bounded by the functions f(x) = x^(2/3) and g(x) = 2x in the first quadrant between y = 2 and y = 3. If Q is rotated around the y-axis, what is the volume of the resulting solid?

Learn how to find the exact volume of the solid formed by rotating the region bounded by f(x) = x^(2/3) and g(x) = 2x in the first quadrant around the y-axis. This detailed solution involves the cylindrical shells method, definite integration, and step-by-step computation. Suitable for undergraduate calculus students.

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