Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. y=^4sqrt(x-1), y=0, x=3
Set up a definite integral that gives the volume, then find the volume of the solid rounded to 4 decimal places.

To find the volume of the solid obtained by rotating the region bounded by the curves about the x-axis, we can use the **disk method**. This method involves integrating the area of circular disks formed by the function as it is revolved around the x-axis.

### Step 1: Define the boundaries
We are given the following:
- \( y = \sqrt[4]{x - 1} \)
- \( y = 0 \) (this is the x-axis)
- \( x = 3 \)

First, we rewrite the function for \( y \):
\[
y = \sqrt[4]{x - 1} \implies y^4 = x - 1 \implies x = y^4 + 1
\]

The region is bounded by the curve \( y = \sqrt[4]{x - 1} \), the x-axis (\( y = 0 \)), and \( x = 3 \).

### Step 2: Set up the definite integral
To use the disk method, we need the radius of each disk, which is the distance from the x-axis to the curve \( y = \sqrt[4]{x - 1} \). The area of each disk is given by \( \pi y^2 \), and the volume is obtained by integrating this area across the interval along the x-axis.

The bounds of the integration will be from \( x = 1 \) to \( x = 3 \) because the curve \( y = \sqrt[4]{x - 1} \) is defined for \( x \geq 1 \).

The volume \( V \) is then given by the integral:
\[
V = \pi \int_{1}^{3} \left( \sqrt[4]{x - 1} \right)^2 \, dx
\]

Since \( \left( \sqrt[4]{x - 1} \right)^2 = (x - 1)^{1/2} \), the integral simplifies to:
\[
V = \pi \int_{1}^{3} (x - 1)^{1/2} \, dx
\]

### Step 3: Solve the integral
To solve the integral:
\[
\int (x - 1)^{1/2} \, dx
\]
we use the substitution \( u = x - 1 \), hence \( du = dx \), and the bounds change from \( x = 1 \) to \( x = 3 \) (i.e., \( u = 0 \) to \( u = 2 \)).

Thus, the integral becomes:
\[
V = \pi \int_{0}^{2} u^{1/2} \, du
\]

Now integrate:
\[
\int u^{1/2} \, du = \frac{2}{3} u^{3/2}
\]

Substitute the bounds:
\[
V = \pi \left[ \frac{2}{3} u^{3/2} \right]_{0}^{2}
\]
\[
V = \pi \left( \frac{2}{3} (2^{3/2}) - 0 \right)
\]
\[
V = \pi \left( \frac{2}{3} \cdot 2\sqrt{2} \right)
\]
\[
V = \pi \cdot \frac{4\sqrt{2}}{3}
\]
\[
V = \frac{4\pi \sqrt{2}}{3}
\]

### Step 4: Calculate the volume
Now, we compute the numerical value:
\[
V = \frac{4\pi \sqrt{2}}{3} \approx 5.9202
\]

Thus, the volume of the solid is approximately **5.9202 cubic units**, rounded to four decimal places.

Would you like any additional details or clarifications? Here are 5 related questions:

1. How would the integral change if the region were rotated around the y-axis instead?
2. Can you explain the substitution method used in integration?
3. How would the disk method differ from the shell method in this case?
4. What if the curve was \( y = \sqrt{x - 1} \) instead of \( y = \sqrt[4]{x - 1} \)?
5. How would we set up the integral if the bounds changed to \( x = 4 \)?

**Tip:** When rotating around an axis, visualize whether the shape forms disks or shells, as this determines the integration method to use.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. y=^4sqrt(x-1), y=0, x=3. Set up a definite integral that gives the volume, then find the volume of the solid rounded to 4 decimal places.

Learn how to calculate the volume of the solid formed by rotating the curve y=^4√(x-1) about the x-axis using the disk method. This detailed solution includes setting up the definite integral and solving it step by step, with the final answer rounded to four decimal places. Perfect for Calculus I or AP Calculus AB students.

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