To find the volume of the solid obtained by rotating the region bounded by the curves about the axis $y = -3$, we can use the method of cylindrical shells. The region is bounded by the curves $y = \frac{1}{x^5}$, $y = 0$, $x = 4$, and $x = 6$.

Step 1: Visualizing the problem

• The region lies between $x = 4$ and $x = 6$ along the x-axis.
• The curve $y = \frac{1}{x^5}$ is decreasing and positive, and it represents the upper boundary of the region.
• The rotation is around the line $y = -3$, which is below the region on the y-axis.

Step 2: Formula for volume using cylindrical shells

For cylindrical shells, the formula for the volume is:

$V = 2\pi \int_{a}^{b} \left( \text{radius of the shell} \right) \times \left( \text{height of the shell} \right) \, dx$

Here:

• The radius of the shell is the distance from the curve $y = \frac{1}{x^5}$ to the line $y = -3$, which is $r(x) = \frac{1}{x^5} + 3$.
• The height of the shell is the value of the function $h(x) = \frac{1}{x^5}$.
• The limits of integration are from $x = 4$ to $x = 6$.

Step 3: Setting up the integral

The volume integral is:

$V = 2\pi \int_{4}^{6} \left( \frac{1}{x^5} + 3 \right) \cdot \frac{1}{x^5} \, dx$

Simplifying the integrand:

$V = 2\pi \int_{4}^{6} \left( \frac{1}{x^{10}} + \frac{3}{x^5} \right) \, dx$

Step 4: Computing the integral

We can now compute the integral term by term.

1. First term: $\int_{4}^{6} \frac{1}{x^{10}} \, dx$ $\int \frac{1}{x^{10}} \, dx = \frac{-1}{9x^9}$ Evaluating from $4$ to $6$: $\left[ \frac{-1}{9x^9} \right]_{4}^{6} = \frac{-1}{9(6^9)} - \frac{-1}{9(4^9)}$

2. Second term: $\int_{4}^{6} \frac{3}{x^5} \, dx$ $\int \frac{3}{x^5} \, dx = \frac{-1}{x^4}$ Evaluating from $4$ to $6$: $\left[ \frac{-1}{x^4} \right]_{4}^{6} = \frac{-1}{6^4} - \frac{-1}{4^4}$

Step 5: Final volume calculation

Let's compute the values of these terms for a final result.

I will now calculate the volume.The volume of the solid obtained by rotating the region bounded by the curves $y = \frac{1}{x^5}$, $y = 0$, $x = 4$, and $x = 6$ about the axis $y = -3$ is approximately $0.01477$ cubic units.

Would you like further details on the steps, or do you have any additional questions?

Here are some related questions for further exploration:

1. How would the volume change if we rotated around the x-axis instead?
2. What if the boundary was $x = 3$ to $x = 5$? How would that affect the volume?
3. Could we use the disk or washer method for this problem, and how would it differ?
4. What happens if the rotation axis is shifted further down, say $y = -5$?
5. How does the shape of the solid change if we rotate around a vertical axis like $x = 0$?

Tip: When rotating around axes that are not the x-axis or y-axis, adjusting the radius of the shells or disks is crucial to accurately computing the volume.