Define Q as the region bounded by the function f(x)=x/2 and the line x=1/2 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.

To calculate the volume of the solid obtained by rotating the region \( Q \) around the \( y \)-axis, we will use the **method of cylindrical shells**. This method is appropriate since we are rotating around the vertical axis, and it requires integrating along the \( y \)-axis.

### Step 1: Set up the function
The region \( Q \) is bounded by the line \( x = 1/2 \), the curve \( f(x) = \frac{x}{2} \), and the horizontal lines \( y = 1 \) and \( y = 3 \). The curve \( f(x) = \frac{x}{2} \) can be rewritten as \( x = 2y \), which expresses \( x \) as a function of \( y \).

### Step 2: Limits of integration
Since the region lies between \( y = 1 \) and \( y = 3 \), these will be the bounds of integration for our calculation.

### Step 3: Shell radius and height
For a given \( y \), the radius of the shell is the distance from the \( y \)-axis, which is simply \( x \). The height of the shell is determined by the difference between the boundaries of \( x \), which are the line \( x = 1/2 \) and the curve \( x = 2y \).

The height of the shell is:
\[
h(y) = 1/2 - 2y
\]
The radius of the shell is:
\[
r(y) = y
\]

### Step 4: Volume of the cylindrical shells
The formula for the volume of a solid of revolution using cylindrical shells is:
\[
V = 2\pi \int_{y_1}^{y_2} \text{radius} \times \text{height} \, dy
\]
In this case:
\[
V = 2\pi \int_{1}^{3} y \cdot \left( \frac{1}{2} - 2y \right) \, dy
\]

### Step 5: Perform the integration
We now expand and integrate:
\[
V = 2\pi \int_1^3 \left( \frac{y}{2} - 2y^2 \right) \, dy
\]
First, integrate each term separately:
\[
\int_1^3 \frac{y}{2} \, dy = \frac{1}{2} \left[ \frac{y^2}{2} \right]_1^3 = \frac{1}{2} \left( \frac{9}{2} - \frac{1}{2} \right) = \frac{1}{2} \times \frac{8}{2} = 2
\]
\[
\int_1^3 2y^2 \, dy = 2 \left[ \frac{y^3}{3} \right]_1^3 = 2 \left( \frac{27}{3} - \frac{1}{3} \right) = 2 \times \frac{26}{3} = \frac{52}{3}
\]

### Step 6: Subtract the results
Now subtract the results of the integrals:
\[
V = 2\pi \left( 2 - \frac{52}{3} \right) = 2\pi \times \left( \frac{6}{3} - \frac{52}{3} \right) = 2\pi \times \frac{-46}{3} = -\frac{92\pi}{3}
\]

Since volume must be positive, we take the absolute value:
\[
V = \frac{92\pi}{3}
\]

Thus, the exact volume of the resulting solid is:
\[
\boxed{\frac{92\pi}{3}}
\]

Would you like any further clarification on this method?

Here are some related questions:
1. What is the difference between the disk and shell method for calculating volumes?
2. How do we rotate regions around axes other than the \(y\)-axis?
3. Can the method of cylindrical shells be applied to other curves and regions?
4. How does the choice of axis affect the method used for volume calculations?
5. Can this method be used for solids with non-linear boundaries?

**Tip:** Always carefully check the limits of integration and functions when setting up an integral for rotation problems.

Define Q as the region bounded by the function f(x)=x/2 and the line x=1/2 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?

Calculate the volume of the solid formed by rotating the region bounded by the function f(x)=x/2 and the line x=1/2 between y=1 and y=3 around the y-axis. This problem applies the cylindrical shell method and key calculus concepts, suitable for undergraduate calculus students.

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