Set up the integrals that would be used to find the volume of the solid generated when the
region bounded by the curves: y = x1/3. y = 0, x=0, and x = 8 using horizontal rectangles when region is revolved around the lines
a. y=0
b. x=0
c. y=10
d. x= -4

To find the volume of the solid generated by revolving the region bounded by the curves \( y = x^{1/3} \), \( y = 0 \), \( x = 0 \), and \( x = 8 \) around different lines, we will use the **disk/washer method** or the **shell method**, depending on the axis of revolution.

### Part (a): Revolving around the line \( y = 0 \) (the x-axis)

1. **Revolution about the x-axis (horizontal axis)** calls for the **disk method** with vertical slices (horizontal rectangles).

The volume integral is:
\[
V = \pi \int_{0}^{8} \left( y_{\text{top}} \right)^2 \, dx
\]
where \( y_{\text{top}} = x^{1/3} \).

Thus, the integral becomes:
\[
V = \pi \int_{0}^{8} \left( x^{1/3} \right)^2 \, dx = \pi \int_{0}^{8} x^{2/3} \, dx
\]

### Part (b): Revolving around the line \( x = 0 \) (the y-axis)

1. **Revolution about the y-axis (vertical axis)** uses the **shell method** with horizontal rectangles.
   - We need to express \( x \) in terms of \( y \): \( x = y^3 \).
   - The bounds for \( y \) are from \( y = 0 \) to \( y = 2 \) (since \( x^{1/3} = 2 \) when \( x = 8 \)).

The volume integral is:
\[
V = 2\pi \int_{0}^{2} (y)(x_{\text{right}}) \, dy
\]
where \( x_{\text{right}} = y^3 \).

Thus, the integral becomes:
\[
V = 2\pi \int_{0}^{2} y \cdot y^3 \, dy = 2\pi \int_{0}^{2} y^4 \, dy
\]

### Part (c): Revolving around the line \( y = 10 \) (a horizontal axis above the curve)

1. This is similar to part (a), but we need to account for the distance between the curve and the axis of revolution. We use the **washer method** with vertical slices (horizontal rectangles).
   - The outer radius is the distance from \( y = 10 \) to \( y = 0 \), which is 10.
   - The inner radius is the distance from \( y = 10 \) to \( y = x^{1/3} \), which is \( 10 - x^{1/3} \).

The volume integral is:
\[
V = \pi \int_{0}^{8} \left( 10^2 - \left( 10 - x^{1/3} \right)^2 \right) dx
\]

### Part (d): Revolving around the line \( x = -4 \) (a vertical axis to the left of the curve)

1. We will use the **shell method** for this, with horizontal rectangles. The distance from \( x \) to the axis \( x = -4 \) is \( x + 4 \).
   - Again, we express \( x \) in terms of \( y \), which is \( x = y^3 \).
   - The bounds for \( y \) are from 0 to 2.

The volume integral is:
\[
V = 2\pi \int_{0}^{2} \left( y^3 + 4 \right) y \, dy
\]

---

Would you like more details or explanations on any of the integrals above?

Here are 5 related questions to further explore this topic:

1. How do we determine when to use the shell method versus the disk/washer method?
2. Can you solve the integrals set up for the volume of revolution for each part?
3. What happens if the region is revolved around a line like \( y = -2 \) instead of \( y = 0 \)?
4. How would the bounds change if the region extended beyond \( x = 8 \)?
5. How does the choice of the axis of rotation affect the complexity of the volume calculation?

**Tip:** When revolving around a vertical or horizontal axis, visualize the slices (rectangles or shells) to choose the right method (disk/washer or shell).

Set up the integrals that would be used to find the volume of the solid generated when the region bounded by the curves: y = x^(1/3), y = 0, x = 0, and x = 8 using horizontal rectangles when region is revolved around the lines a. y = 0 b. x = 0 c. y = 10 d. x = -4

Learn how to set up integrals for finding the volume of solids generated by revolving the region bounded by y = x^(1/3), y = 0, x = 0, and x = 8 around various lines using the disk and shell methods. This guide covers key calculus concepts including volumes of revolution.

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