Using the disk method, determine the volume of a solid formed by revolving the region bounded above by the line y=1, on the left by the line x=0, and below by the curve the x=y^3/2 about the y=1 line . using disc method 
Setup and solve the integral that represents the volume of the solid of revolution described above 

To solve this problem using the **disk method**, we need to set up the integral for the volume of the solid formed by revolving the region about the line \( y = 1 \).

### 1. **Identify the boundaries and curve:**
- The region is bounded:
  - Above by the line \( y = 1 \)
  - On the left by the line \( x = 0 \)
  - Below by the curve \( x = y^{3/2} \)

### 2. **Transformation for the disk method:**
Since we are revolving around the horizontal line \( y = 1 \), the radius of each disk will be the distance between the curve and the line \( y = 1 \).

The radius of a disk at a particular \( y \)-coordinate is the vertical distance between the line \( y = 1 \) and the curve \( x = y^{3/2} \). Since we are using horizontal disks, the radius is given by:

\[
r(y) = 1 - y
\]

The cross-sectional area of each disk will be:

\[
A(y) = \pi [r(y)]^2 = \pi (1 - y)^2
\]

We are revolving over the interval of \( y \), which is from \( y = 0 \) to \( y = 1 \).

### 3. **Set up the volume integral:**

The formula for the volume using the disk method is:

\[
V = \int_a^b \pi [r(y)]^2 \, dy
\]

For our problem, this becomes:

\[
V = \int_0^1 \pi (1 - y)^2 \, dy
\]

### 4. **Solve the integral:**

First, expand the integrand:

\[
(1 - y)^2 = 1 - 2y + y^2
\]

Thus, the volume integral becomes:

\[
V = \pi \int_0^1 (1 - 2y + y^2) \, dy
\]

Now, integrate term by term:

\[
V = \pi \left[ \int_0^1 1 \, dy - 2 \int_0^1 y \, dy + \int_0^1 y^2 \, dy \right]
\]

Calculate each term:

\[
\int_0^1 1 \, dy = y \Big|_0^1 = 1
\]
\[
\int_0^1 y \, dy = \frac{y^2}{2} \Big|_0^1 = \frac{1}{2}
\]
\[
\int_0^1 y^2 \, dy = \frac{y^3}{3} \Big|_0^1 = \frac{1}{3}
\]

Thus, the integral becomes:

\[
V = \pi \left[ 1 - 2 \times \frac{1}{2} + \frac{1}{3} \right] = \pi \left[ 1 - 1 + \frac{1}{3} \right] = \pi \times \frac{1}{3}
\]

### 5. **Final answer:**

The volume of the solid is:

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

---

Would you like further details or have any questions about this solution?

### Related questions:
1. How would the integral change if the solid were rotated around the x-axis instead?
2. What if the region were rotated around the line \( y = 2 \) instead of \( y = 1 \)?
3. Can you apply the shell method to this problem for comparison?
4. What is the effect of changing the curve to \( x = y^2 \) on the volume?
5. How do we handle the volume if the region had different bounds on \( x \)?

### Tip:
When using the disk method, always ensure the radius is measured as the distance from the axis of rotation to the curve.

Using the disk method, determine the volume of a solid formed by revolving the region bounded above by the line y=1, on the left by the line x=0, and below by the curve x=y^3/2 about the y=1 line. Setup and solve the integral that represents the volume of the solid of revolution described above.

Explore how to calculate the volume of a solid formed by revolving the region bounded by y=1, x=0, and the curve x=y^(3/2) around the line y=1. This detailed solution applies the disk method and integral calculus, ideal for advanced high school or college calculus students.

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