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:

$V = \frac{\pi}{3}$

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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.