Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.
y = x5,    y = 

squareroot(x)
;    about the x-axis

To find the volume of the solid obtained by rotating the region bounded by the curves \(y = x^5\) and \(y = \sqrt{x}\) about the x-axis, we'll use the disk/washer method. Here’s a step-by-step approach:

### Step 1: Sketch the Region
First, let's visualize the region:

1. **Curves**: 
   - \(y = x^5\) is a steep curve that increases slowly for small \(x\) and becomes steeper as \(x\) increases.
   - \(y = \sqrt{x}\) increases rapidly at the beginning and levels off as \(x\) increases.
   
2. **Intersection Points**:
   - Find where the curves intersect by setting \(x^5 = \sqrt{x}\).
   
   Simplifying this:
   \[
   x^5 = x^{1/2}
   \]
   \[
   x^{10} = x
   \]
   \[
   x^{10} - x = 0
   \]
   \[
   x(x^9 - 1) = 0
   \]
   The solutions are \(x = 0\) and \(x = 1\).

   So, the curves intersect at \(x = 0\) and \(x = 1\).

### Step 2: Set Up the Integral Using the Washer Method
Since we are rotating around the x-axis, the volume can be computed using the washer method. A washer has an outer radius (top curve) and an inner radius (bottom curve).

For \(x\) from \(0\) to \(1\):
- **Outer radius**: \(R(x) = \sqrt{x}\)
- **Inner radius**: \(r(x) = x^5\)

The volume \(V\) of the solid of revolution is given by:
\[
V = \pi \int_{a}^{b} \left(R(x)^2 - r(x)^2\right) \, dx
\]
Substituting our values:
\[
V = \pi \int_{0}^{1} \left((\sqrt{x})^2 - (x^5)^2\right) \, dx
\]
Simplifying inside the integral:
\[
V = \pi \int_{0}^{1} \left(x - x^{10}\right) \, dx
\]

### Step 3: Evaluate the Integral
Now, integrate with respect to \(x\):
\[
V = \pi \left[ \frac{x^2}{2} - \frac{x^{11}}{11} \right]_{0}^{1}
\]
Evaluating at the bounds:
\[
V = \pi \left( \frac{1^2}{2} - \frac{1^{11}}{11} \right) - \pi \left( \frac{0^2}{2} - \frac{0^{11}}{11} \right)
\]
\[
V = \pi \left( \frac{1}{2} - \frac{1}{11} \right)
\]
\[
V = \pi \left( \frac{11}{22} - \frac{2}{22} \right)
\]
\[
V = \pi \left( \frac{9}{22} \right)
\]
\[
V = \frac{9\pi}{22}
\]

Thus, the volume of the solid is \(\frac{9\pi}{22}\).

Would you like a detailed sketch of the region, the solid, and a typical washer?

### Related Questions:
1. How would the volume change if the solid was rotated about the y-axis instead?
2. What if the curves were rotated about the line \(y = 1\)? How would the setup change?
3. What happens to the volume if the region is extended to \(x = 2\)?
4. How does the disk method differ when the region is rotated around the x-axis versus the y-axis?
5. Can you calculate the surface area of the solid formed by rotating the same region?

**Tip:** When setting up integrals for volumes of solids of revolution, carefully identify the outer and inner radii to ensure correct calculation of the volume.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. y = x^5, y = sqrt(x); about the x-axis

Learn how to calculate the volume of the solid formed by rotating the region bounded by y = x^5 and y = sqrt(x) around the x-axis. This step-by-step guide uses the washer method and covers key calculus concepts such as integration and the fundamental theorem of calculus.

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