I have a triangle bound by y=2x, y=9, and x=0.5, what is the integral of the solid rotated around the y-axis *use dy not dx

The region of the triangle is bounded by the lines \( y = 2x \), \( y = 9 \), and \( x = 0.5 \), and we are rotating this region around the y-axis. To calculate the volume of the solid formed by this rotation using the method of cylindrical shells, we will set up the integral with respect to \( y \) (since we are using \( dy \)).

### Step-by-step process:

1. **Equation of the boundary lines**:
    - \( y = 2x \) can be rewritten as \( x = \frac{y}{2} \).
    - The vertical line is \( x = 0.5 \).
    - The horizontal line is \( y = 9 \).

2. **Set up the integral using cylindrical shells**:
    The formula for the volume using the shell method when rotating around the \( y \)-axis is:
    \[
    V = 2\pi \int_{y_{\text{min}}}^{y_{\text{max}}} (x \cdot \text{height}) \, dy
    \]
    where \( x \) is the distance from the axis of rotation (the y-axis), and the height of each shell is the corresponding width of the triangle at a given \( y \).

    - The height (or the width of the triangle at a given \( y \)) is \( 0.5 - \frac{y}{2} \), because at a given \( y \), the right boundary of the triangle is at \( x = 0.5 \) and the left boundary is \( x = \frac{y}{2} \).
  
3. **Integral bounds**:
    - The lower bound for \( y \) is at \( y = 0 \) (where the triangle starts).
    - The upper bound is at \( y = 9 \) (where the horizontal line \( y = 9 \) intersects the region).

4. **Set up the integral**:
    The volume \( V \) is given by:
    \[
    V = 2\pi \int_0^9 \left( \left( 0.5 \right) - \frac{y}{2} \right) \, dy
    \]

5. **Evaluate the integral**:
    \[
    V = 2\pi \left[ 0.5y - \frac{y^2}{4} \right]_0^9
    \]

    Substitute the bounds into the expression:

    \[
    V = 2\pi \left( \left[ 0.5(9) - \frac{9^2}{4} \right] - \left[ 0.5(0) - \frac{0^2}{4} \right] \right)
    \]
    \[
    V = 2\pi \left( 4.5 - \frac{81}{4} \right)
    \]
    \[
    V = 2\pi \left( 4.5 - 20.25 \right)
    \]
    \[
    V = 2\pi (-15.75)
    \]
    \[
    V = -31.5\pi
    \]

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

\[
V = 31.5\pi \, \text{units}^3
\]

Thus, the volume of the solid generated by rotating the region around the \( y \)-axis is \( 31.5\pi \, \text{units}^3 \).

Would you like more details on any step or clarifications?

Here are 5 related questions:
1. What if the rotation was around the x-axis instead?
2. How does the method of cylindrical shells differ from the disk/washer method?
3. What is the significance of integrating with respect to \( dy \) versus \( dx \)?
4. How would you handle different types of regions being rotated around the y-axis?
5. Can you explain the physical interpretation of the volume formula for cylindrical shells?

**Tip:** When integrating for volumes, always ensure that the limits and variable of integration match the axis of rotation.

Learn how to calculate the volume of a solid formed by rotating a triangular region around the y-axis using the method of cylindrical shells. This problem involves geometry, integral calculus, and the formula for volumes of revolution. Suitable for undergraduate calculus students.

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