Find the volume formed by rotating about the y-axis the region enclosed by: x=8y and y^3=x with y≥0

We are asked to find the volume of the solid formed by rotating the region enclosed by the curves \( x = 8y \) and \( y^3 = x \) about the y-axis, with the restriction \( y \geq 0 \).

### Step 1: Express the equations
The given equations are:
1. \( x = 8y \)
2. \( y^3 = x \)

We are rotating around the y-axis, so we need to express the curves in terms of \( y \).

For the first equation:
\[
x = 8y
\]

For the second equation:
\[
x = y^3
\]

### Step 2: Find the points of intersection
To determine the limits of integration, we need to find where the two curves intersect. Set the two equations equal to each other:
\[
8y = y^3
\]

Rearrange and solve for \( y \):
\[
y^3 - 8y = 0
\]
\[
y(y^2 - 8) = 0
\]
\[
y = 0 \quad \text{or} \quad y^2 = 8 \quad \Rightarrow \quad y = \sqrt{8} = 2\sqrt{2}
\]

Thus, the curves intersect at \( y = 0 \) and \( y = 2\sqrt{2} \).

### Step 3: Set up the volume integral using the method of cylindrical shells
Since we are rotating around the y-axis, we will use the method of cylindrical shells. The formula for the volume using cylindrical shells is:
\[
V = 2\pi \int_{a}^{b} \text{radius} \times \text{height} \, dy
\]
Here:
- The radius is \( y \) (distance from the y-axis).
- The height is the difference between the two functions \( x = 8y \) and \( x = y^3 \), i.e., \( 8y - y^3 \).

Thus, the volume integral becomes:
\[
V = 2\pi \int_{0}^{2\sqrt{2}} y(8y - y^3) \, dy
\]

### Step 4: Simplify and integrate
First, simplify the integrand:
\[
y(8y - y^3) = 8y^2 - y^4
\]

Now, set up the integral:
\[
V = 2\pi \int_{0}^{2\sqrt{2}} (8y^2 - y^4) \, dy
\]

Integrate term by term:
\[
\int 8y^2 \, dy = \frac{8y^3}{3}
\]
\[
\int y^4 \, dy = \frac{y^5}{5}
\]

Now evaluate the definite integrals:
\[
V = 2\pi \left[ \frac{8y^3}{3} - \frac{y^5}{5} \right]_{0}^{2\sqrt{2}}
\]

### Step 5: Compute the values at the limits
First, calculate \( \left( 2\sqrt{2} \right)^3 \) and \( \left( 2\sqrt{2} \right)^5 \):
\[
\left( 2\sqrt{2} \right)^3 = 16\sqrt{2}, \quad \left( 2\sqrt{2} \right)^5 = 64\sqrt{2}
\]

Now substitute into the expression:
\[
V = 2\pi \left( \frac{8 \cdot 16\sqrt{2}}{3} - \frac{64\sqrt{2}}{5} \right)
\]
\[
V = 2\pi \left( \frac{128\sqrt{2}}{3} - \frac{64\sqrt{2}}{5} \right)
\]

### Step 6: Combine the terms
Find a common denominator:
\[
V = 2\pi \left( \frac{128\sqrt{2} \cdot 5}{15} - \frac{64\sqrt{2} \cdot 3}{15} \right)
\]
\[
V = 2\pi \left( \frac{640\sqrt{2} - 192\sqrt{2}}{15} \right)
\]
\[
V = 2\pi \cdot \frac{448\sqrt{2}}{15}
\]

Finally, simplify the result:
\[
V = \frac{896\pi\sqrt{2}}{15}
\]

Thus, the volume of the solid is \( \frac{896\pi\sqrt{2}}{15} \).

Would you like more details or further clarifications?

### Related Questions:
1. What is the difference between using the disk method and the shell method?
2. How do you find the points of intersection between two curves analytically?
3. What is the geometrical meaning of the volume generated by rotating a region?
4. How does rotating around different axes affect the integral setup?
5. Can the same method be applied if we rotate around the x-axis instead of the y-axis?

### Tip:
Always sketch the region and the axis of rotation to visualize the problem better. This helps in choosing the correct method and setting up the integral properly.

Find the volume formed by rotating about the y-axis the region enclosed by: x=8y and y^3=x with y≥0.

Learn how to calculate the volume of the solid formed by rotating the region enclosed by x=8y and y^3=x about the y-axis, using the method of cylindrical shells. The solution includes finding the points of intersection, setting up the integral, and evaluating the volume. Suitable for first-year calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation