Find the volume formed by rotating the region enclosed by:

x=4y, y^3=x with y≥0

about the y-axis

volume=

We are tasked with finding the volume of the solid formed by rotating the region enclosed by the curves \( x = 4y \) and \( y^3 = x \), where \( y \geq 0 \), about the y-axis.

### Step 1: Determine the points of intersection
To find the region, we first need to determine where the two curves intersect by setting \( x = 4y \) equal to \( x = y^3 \):
\[
4y = y^3
\]
Dividing both sides by \( y \) (since \( y \geq 0 \) excludes \( y = 0 \)):
\[
4 = y^2
\]
Thus, \( y = 2 \).

The curves intersect at \( y = 0 \) and \( y = 2 \).

### Step 2: Set up the volume integral
To compute the volume formed by rotating around the y-axis, we use the method of cylindrical shells. The formula for the volume of a solid formed by rotating about the y-axis is given by:
\[
V = 2\pi \int_{a}^{b} x \cdot f(y) \, dy
\]
where \( x \) is the radius and \( f(y) \) is the function for the height of the shell.

- For \( y^3 = x \), the radius is \( x = y^3 \).
- For \( x = 4y \), the radius is \( x = 4y \).

We will subtract the inner radius (represented by \( y^3 \)) from the outer radius (represented by \( 4y \)) over the interval \( y = 0 \) to \( y = 2 \).

The volume is:
\[
V = 2\pi \int_0^2 (4y - y^3) \, y \, dy
\]
This simplifies to:
\[
V = 2\pi \int_0^2 (4y^2 - y^4) \, dy
\]

### Step 3: Integrate
Now, compute the integral:
\[
\int_0^2 (4y^2 - y^4) \, dy = \left[\frac{4y^3}{3} - \frac{y^5}{5}\right]_0^2
\]
Evaluating at the bounds:
\[
\left(\frac{4(2)^3}{3} - \frac{(2)^5}{5}\right) - \left(\frac{4(0)^3}{3} - \frac{(0)^5}{5}\right)
= \left(\frac{4(8)}{3} - \frac{32}{5}\right)
= \left(\frac{32}{3} - \frac{32}{5}\right)
\]
Now, find a common denominator and subtract:
\[
\frac{32}{3} - \frac{32}{5} = \frac{160}{15} - \frac{96}{15} = \frac{64}{15}
\]

### Step 4: Multiply by \( 2\pi \)
Finally, multiply by \( 2\pi \) to get the volume:
\[
V = 2\pi \cdot \frac{64}{15} = \frac{128\pi}{15}
\]

### Final Answer:
The volume of the solid formed by rotating the region about the y-axis is:
\[
V = \frac{128\pi}{15}
\]

Would you like more details on any part of this process?

---

Here are 5 related questions:
1. How would the volume change if the region was rotated about the x-axis instead?
2. Can you compute the surface area of the same solid of revolution?
3. How would you calculate the volume if the curves were \( x = 4y \) and \( y^2 = x \)?
4. What method could be used to calculate the volume if the curves enclosed a different region?
5. How does the method of disks differ from the method of cylindrical shells?

**Tip:** When solving volume problems involving rotation, always visualize the region and identify the axis of rotation carefully before setting up the integral.

Find the volume formed by rotating the region enclosed by: x = 4y, y^3 = x with y ≥ 0 about the y-axis

This problem involves calculating the volume of a solid formed by rotating the region enclosed by the curves x = 4y and y^3 = x about the y-axis using the cylindrical shell method. The solution covers calculus concepts including integration and the fundamental theorem of calculus, suitable for students in Grades 11-12 or early college.

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