We need to find the volume of the solid formed by rotating the region enclosed by:

$x = 0, \quad x = 1, \quad y = 0, \quad y = 3 + x^8$

about the y-axis.

Step 1: Choose the Shell Method

Since we are rotating around the y-axis and the given boundaries are in terms of $x$, the shell method is the best choice.

The shell volume formula is:

$V = \int_{a}^{b} 2\pi (\text{radius}) (\text{height}) \, dx$

• Radius: Since we are rotating around the y-axis, the radius is simply $x$.
• Height: The height is the function $y = 3 + x^8$.

Step 2: Set Up the Integral

The limits are from $x = 0$ to $x = 1$:

$V = \int_0^1 2\pi x (3 + x^8) \, dx$

Step 3: Compute the Integral

Expanding:

$V = 2\pi \int_0^1 (3x + x^9) \, dx$

Now, integrating term by term:

$2\pi \left[ \frac{3x^2}{2} + \frac{x^{10}}{10} \right] \Bigg|_0^1$

Evaluating at $x = 1$:

[ 2\pi \left[ \frac{3(1)^2}{2} + \frac{(1)^{10}}{10} \right] = 2\pi \left[ \frac{3}{2} + \frac{1}{10} \right] ]

Convert to a common denominator (10):

$\frac{3}{2} = \frac{15}{10}, \quad \frac{1}{10} = \frac{1}{10}$

$\frac{15}{10} + \frac{1}{10} = \frac{16}{10} = \frac{8}{5}$

Step 4: Final Answer

$V = 2\pi \times \frac{8}{5} = \frac{16\pi}{5}$

Final Answer:

$\frac{16\pi}{5}$

Would you like a deeper explanation or a different approach?

Related Questions:

1. How would the integral change if the rotation was around the $x$-axis instead?
2. What if the function was $y = 3 - x^8$ instead of $y = 3 + x^8$?
3. How do you determine whether to use the shell method or the disk/washer method?
4. What happens if the region is rotated about $x = 2$ instead of the $y$-axis?
5. Can we use cylindrical coordinates to solve this problem?

Tip:

The shell method is most useful when integrating along the x-axis but rotating around the y-axis (or vice versa). Always check whether it simplifies your integral.