Define Q as the region bounded by the functions f(x)=x ^2/3 and g(x)=1/x in the first quadrant between y=1 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.

We are given two functions \( f(x) = x^{2/3} \) and \( g(x) = \frac{1}{x} \), and the region \( Q \) is bounded between \( y = 1 \) and \( y = 3 \) in the first quadrant. The volume of the solid formed by rotating this region around the \( y \)-axis can be calculated using the method of cylindrical shells.

### Step 1: Solve for \( x \) in terms of \( y \)
We need to express \( x \) as a function of \( y \) for both \( f(x) \) and \( g(x) \), since the region is rotated around the \( y \)-axis, and we want to integrate with respect to \( y \).

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

For \( g(x) = \frac{1}{x} \):
\[
y = \frac{1}{x} \implies x = \frac{1}{y}.
\]

### Step 2: Set up the volume integral
The volume of the solid generated by rotating the region around the \( y \)-axis can be calculated using the formula for the volume of a solid of revolution:

\[
V = \pi \int_{y_1}^{y_2} \left[ R_{\text{outer}}(y)^2 - R_{\text{inner}}(y)^2 \right] \, dy,
\]
where:
- \( R_{\text{outer}}(y) \) is the outer radius of the shell, corresponding to the function \( x = \frac{1}{y} \),
- \( R_{\text{inner}}(y) \) is the inner radius of the shell, corresponding to the function \( x = y^{3/2} \),
- The bounds are from \( y_1 = 1 \) to \( y_2 = 3 \).

Thus, the volume integral is:

\[
V = \pi \int_1^3 \left( \left( \frac{1}{y} \right)^2 - \left( y^{3/2} \right)^2 \right) \, dy.
\]

### Step 3: Simplify the integrand
Now, simplify the expressions inside the integral:

\[
\left( \frac{1}{y} \right)^2 = \frac{1}{y^2}, \quad \left( y^{3/2} \right)^2 = y^3.
\]

So the volume integral becomes:

\[
V = \pi \int_1^3 \left( \frac{1}{y^2} - y^3 \right) \, dy.
\]

### Step 4: Evaluate the integral
We now integrate each term separately:

\[
\int_1^3 \frac{1}{y^2} \, dy = \left[ -\frac{1}{y} \right]_1^3 = -\frac{1}{3} + 1 = \frac{2}{3},
\]
and
\[
\int_1^3 y^3 \, dy = \left[ \frac{y^4}{4} \right]_1^3 = \frac{3^4}{4} - \frac{1^4}{4} = \frac{81}{4} - \frac{1}{4} = \frac{80}{4} = 20.
\]

### Step 5: Combine the results
Now, subtract the two integrals:

\[
V = \pi \left( \frac{2}{3} - 20 \right) = \pi \left( \frac{2}{3} - \frac{60}{3} \right) = \pi \left( \frac{2 - 60}{3} \right) = \pi \left( \frac{-58}{3} \right).
\]

Thus, the volume of the resulting solid is:

\[
V = -\frac{58\pi}{3}.
\]

Since volume cannot be negative, we take the absolute value:

\[
V = \frac{58\pi}{3}.
\]

### Final Answer:
The volume of the resulting solid is \( \frac{58\pi}{3} \).

Would you like more details on this solution? Here are five related questions for further exploration:

1. How do we derive the formula for the volume of a solid of revolution using cylindrical shells?
2. What are other methods to calculate the volume of a solid, such as the disk or washer method?
3. How does the choice of axis of rotation affect the resulting volume?
4. What happens to the volume if the region is rotated around a different line, like \( y = 1 \)?
5. How do you compute the volume of a solid when the functions involved are more complex?

**Tip:** When setting up integrals for volumes of revolution, always carefully distinguish between outer and inner radii.

Define Q as the region bounded by the functions f(x)=x ^2/3 and g(x)=1/x in the first quadrant between y=1 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.

Find the volume of a solid formed by rotating the region bounded by f(x) = x^(2/3) and g(x) = 1/x in the first quadrant between y = 1 and y = 3 around the y-axis. This calculus problem covers the cylindrical shell method and integral calculus, offering a step-by-step solution. Ideal for college-level students.

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