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The problem in the image involves finding the volume of a solid generated by revolving the region bounded by the curve \( y = x^3 \) and \( y = x^6 \) about the y-axis.

### Step 1: Set up the integral

The curves given are:
\[
y = x^3
\]
\[
y = x^6
\]

When the region bounded by these curves is revolved around the y-axis, we use the method of cylindrical shells. The volume \( V \) is given by the integral:
\[
V = 2\pi \int_{a}^{b} x \cdot (f(x) - g(x)) \, dx
\]
where \( f(x) \) and \( g(x) \) are the outer and inner functions, respectively. In this case, \( f(x) = x^3 \) and \( g(x) = x^6 \).

### Step 2: Find the intersection points

To find the limits of integration, set the curves equal to each other:
\[
x^3 = x^6
\]
\[
x^3(1 - x^3) = 0
\]
So, \( x = 0 \) or \( x = 1 \).

Thus, the limits of integration are from \( x = 0 \) to \( x = 1 \).

### Step 3: Compute the integral

The volume is:
\[
V = 2\pi \int_{0}^{1} x \cdot (x^3 - x^6) \, dx
\]
\[
V = 2\pi \int_{0}^{1} (x^4 - x^7) \, dx
\]

### Step 4: Evaluate the integral

Now, integrate term by term:
\[
\int x^4 \, dx = \frac{x^5}{5}
\]
\[
\int x^7 \, dx = \frac{x^8}{8}
\]

So, the volume integral becomes:
\[
V = 2\pi \left[\frac{x^5}{5} - \frac{x^8}{8}\right]_{0}^{1}
\]

### Step 5: Calculate the definite integral

Substitute the limits into the antiderivative:
\[
V = 2\pi \left(\left[\frac{1^5}{5} - \frac{1^8}{8}\right] - \left[\frac{0^5}{5} - \frac{0^8}{8}\right]\right)
\]
\[
V = 2\pi \left(\frac{1}{5} - \frac{1}{8}\right)
\]

### Step 6: Simplify

To combine the fractions:
\[
\frac{1}{5} - \frac{1}{8} = \frac{8 - 5}{40} = \frac{3}{40}
\]

Thus, the volume is:
\[
V = 2\pi \cdot \frac{3}{40} = \frac{6\pi}{40} = \frac{3\pi}{20}
\]

So the final answer is:
\[
\boxed{\frac{3\pi}{20}}
\]

Would you like to explore any more details or have any questions?

### Related Questions:
1. What would the volume be if the region were rotated around the x-axis instead of the y-axis?
2. How would the problem change if the bounds of integration were different, such as from \( x = 0 \) to \( x = 2 \)?
3. Can you find the area between the curves \( y = x^3 \) and \( y = x^6 \) without rotation?
4. What is the geometric interpretation of the volume found?
5. How does the method of cylindrical shells compare to the disk/washer method for this problem?

### Tip:
When solving volumes of revolution, always carefully consider the axis of rotation, as it determines the correct method (disk/washer or cylindrical shells) and the setup of your integral.

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the y-axis.
y = x^3, y = x^6

Learn how to calculate the volume of a solid generated by revolving the region bounded by y = x^3 and y = x^6 about the y-axis. This problem involves key calculus concepts, including the method of cylindrical shells and definite integral evaluation. Suitable for students in Grades 11-12.

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