Let R be the region bounded by the following curves: y = x^(1/3),x = 8,y = 0.Set up the integration to find the volume of the solid generated by revolving R around: the x axis

To find the volume of the solid generated by revolving the region \( R \) around the x-axis, we will use the **disk method**. The region \( R \) is bounded by the curves \( y = x^{1/3} \), \( x = 8 \), and \( y = 0 \). Here’s the step-by-step process for setting up the integral:

### 1. The Disk Method Formula:
The volume of a solid generated by revolving a region around the x-axis is given by the formula:
\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]
where \( f(x) \) represents the function defining the curve, and the limits of integration are from \( a \) to \( b \) along the x-axis.

### 2. Identifying the function and limits:
- The curve is \( y = x^{1/3} \), which can be written as \( f(x) = x^{1/3} \).
- The region is bounded by \( x = 0 \) and \( x = 8 \), as \( y = 0 \) corresponds to the x-axis and the upper boundary is \( x = 8 \).

### 3. Setting up the integral:
Now, we apply the disk method formula with \( f(x) = x^{1/3} \) and the limits of integration from \( 0 \) to \( 8 \):
\[
V = \pi \int_{0}^{8} \left( x^{1/3} \right)^2 \, dx
\]

### 4. Simplifying the integrand:
\[
V = \pi \int_{0}^{8} x^{2/3} \, dx
\]

### 5. Performing the integration:
The integral of \( x^{2/3} \) is:
\[
\int x^{2/3} \, dx = \frac{3}{5} x^{5/3}
\]

Thus, the volume becomes:
\[
V = \pi \left[ \frac{3}{5} x^{5/3} \right]_{0}^{8}
\]

### 6. Evaluate the definite integral:
Now, substitute the limits of integration:
\[
V = \pi \left( \frac{3}{5} \times 8^{5/3} - \frac{3}{5} \times 0^{5/3} \right)
\]
Since \( 0^{5/3} = 0 \), we only need to evaluate \( 8^{5/3} \):
\[
8^{5/3} = (2^3)^{5/3} = 2^5 = 32
\]
So, the volume is:
\[
V = \pi \times \frac{3}{5} \times 32 = \frac{96\pi}{5}
\]

Thus, the volume of the solid is:
\[
V = \frac{96\pi}{5}
\]

### Final Answer:
The volume of the solid generated by revolving the region around the x-axis is \( \frac{96\pi}{5} \) cubic units.

---

Would you like further details on this method or have any other questions?

Here are 5 related questions for further exploration:
1. How does the method change if we revolve around the y-axis instead?
2. What if the region were bounded by \( y = x^{1/2} \) instead of \( y = x^{1/3} \)?
3. How do you calculate volume using the shell method instead of the disk method?
4. How does the presence of additional boundaries, such as \( y = 1 \), affect the volume calculation?
5. Can this problem be solved using polar coordinates, and how?

**Tip:** The disk method works well when revolving around horizontal or vertical lines. If the axis of revolution is not the x- or y-axis, consider using the **washer** or **shell method**.

Let R be the region bounded by the following curves: y = x^(1/3), x = 8, y = 0. Set up the integration to find the volume of the solid generated by revolving R around the x-axis.

Learn how to calculate the volume of a solid generated by revolving the region bounded by y = x^(1/3), x = 8, and y = 0 around the x-axis. This detailed solution uses the disk method and is suitable for students in Grades 11-12 studying advanced calculus.

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