You want to find the volume of the solid obtained by rotating about the x-axis the region under the curve y=x1/5𝑦=𝑥1/5 from 0 to 2.

You first slice through the rotated solid at a generic point x𝑥 and get a circular cross-section. What is the area of the circular cross-section?

A(x)=𝐴(𝑥)=πx25​Correct  

This makes the volume of the approximating disk with thickness ΔxΔ𝑥 equal to which expression?

Volume of disk = πx25​Correct ΔxΔ𝑥  

Now let ΔxΔ𝑥 approach 0, and sum the volumes of the infinitely many disks that approximate the solid of revolution. What total volume do you get?

V=∫20A(x)dx=

To find the total volume \( V \) of the solid obtained by rotating the region under the curve \( y = x^{1/5} \) around the x-axis from \( x = 0 \) to \( x = 2 \), we follow these steps:

### Step 1: Area of the circular cross-section
The area of a circular cross-section at a generic point \( x \) is given by:
\[
A(x) = \pi \left( y(x) \right)^2 = \pi \left( x^{1/5} \right)^2 = \pi x^{2/5}.
\]

### Step 2: Volume of an approximating disk
The volume of an approximating disk at a point \( x \) with a small thickness \( \Delta x \) is:
\[
\text{Volume of disk} = A(x) \Delta x = \pi x^{2/5} \Delta x.
\]

### Step 3: Total volume of the solid
To find the total volume, sum up the volumes of the disks as \( \Delta x \) approaches 0. This is done by integrating the area function \( A(x) \) from \( x = 0 \) to \( x = 2 \):
\[
V = \int_0^2 A(x) \, dx = \int_0^2 \pi x^{2/5} \, dx.
\]
We can factor out the constant \( \pi \):
\[
V = \pi \int_0^2 x^{2/5} \, dx.
\]

### Step 4: Compute the integral
To compute \( \int_0^2 x^{2/5} \, dx \), use the power rule of integration:
\[
\int x^n \, dx = \frac{x^{n+1}}{n+1}, \quad \text{where } n = \frac{2}{5}.
\]
So,
\[
\int_0^2 x^{2/5} \, dx = \left[ \frac{x^{7/5}}{7/5} \right]_0^2 = \frac{5}{7} \left( 2^{7/5} - 0^{7/5} \right).
\]
Since \( 0^{7/5} = 0 \), this simplifies to:
\[
\int_0^2 x^{2/5} \, dx = \frac{5}{7} \cdot 2^{7/5}.
\]

### Step 5: Final volume expression
Now, multiplying by \( \pi \):
\[
V = \pi \cdot \frac{5}{7} \cdot 2^{7/5}.
\]

Thus, the total volume of the solid is:
\[
V = \frac{5\pi}{7} \cdot 2^{7/5}.
\]

Would you like a numerical evaluation of this result or more details on any step?

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Here are 5 related questions to expand on this concept:
1. How would the volume change if the curve was rotated about the y-axis instead?
2. What is the volume if the curve is \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 2 \)?
3. How does the integral setup change if we rotate around a horizontal line \( y = c \)?
4. How can we find the surface area of the same solid instead of the volume?
5. What happens if we rotate the region under the curve about the line \( x = 3 \)?

**Tip:** For solids of revolution, always check if the cross-section is perpendicular to the axis of rotation to correctly apply the disk or washer method.

You want to find the volume of the solid obtained by rotating about the x-axis the region under the curve y = x^(1/5) from 0 to 2. You first slice through the rotated solid at a generic point x and get a circular cross-section. What is the area of the circular cross-section? A(x) = πx^(2/5) Correct. This makes the volume of the approximating disk with thickness Δx equal to which expression? Volume of disk = πx^(2/5) Δx. Now let Δx approach 0, and sum the volumes of the infinitely many disks that approximate the solid of revolution. What total volume do you get? V = ∫(0 to 2) A(x) dx.

Discover how to calculate the volume of a solid obtained by rotating the region under the curve y = x^(1/5) around the x-axis from 0 to 2. This problem uses the disk method and the power rule of integration, making it perfect for college-level calculus students.

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