Find the volume of the solid that results from rotating the region bounded by the graphs of y−6x−1=0
, y=0
, and x=5
 about the x
-axis. Write the exact answer. Do not round.


To find the volume of the solid formed by rotating the region bounded by the curves \( y - 6x - 1 = 0 \), \( y = 0 \), and \( x = 5 \) about the \( x \)-axis, we will use the **disk method**.

### Step-by-Step Process:

1. **Solve for \( y \):**
   The equation of the line is:
   \[
   y - 6x - 1 = 0
   \]
   Solving for \( y \), we get:
   \[
   y = 6x + 1
   \]

2. **Set up the disk method formula:**
   The general formula for the volume \( V \) of a solid of revolution using the disk method is:
   \[
   V = \pi \int_{a}^{b} [f(x)]^2 \, dx
   \]
   where \( f(x) \) is the function being rotated around the x-axis, and \( a \) and \( b \) are the bounds of integration.

3. **Define the bounds and function:**
   The region is bounded by \( x = 0 \) and \( x = 5 \) (since the region is bounded by the vertical line \( x = 5 \)), and the function is \( f(x) = 6x + 1 \).

4. **Set up the integral:**
   The volume is given by:
   \[
   V = \pi \int_{0}^{5} (6x + 1)^2 \, dx
   \]

5. **Expand the integrand:**
   First, expand \( (6x + 1)^2 \):
   \[
   (6x + 1)^2 = 36x^2 + 12x + 1
   \]
   So the integral becomes:
   \[
   V = \pi \int_{0}^{5} (36x^2 + 12x + 1) \, dx
   \]

6. **Integrate:**
   Now, integrate each term:
   \[
   \int 36x^2 \, dx = 12x^3
   \]
   \[
   \int 12x \, dx = 6x^2
   \]
   \[
   \int 1 \, dx = x
   \]
   Therefore:
   \[
   V = \pi \left[ 12x^3 + 6x^2 + x \right]_{0}^{5}
   \]

7. **Evaluate the definite integral:**
   At \( x = 5 \):
   \[
   12(5)^3 + 6(5)^2 + (5) = 12(125) + 6(25) + 5 = 1500 + 150 + 5 = 1655
   \]
   At \( x = 0 \):
   \[
   12(0)^3 + 6(0)^2 + (0) = 0
   \]
   So, the volume is:
   \[
   V = \pi (1655 - 0) = 1655\pi
   \]

### Final Answer:
The exact volume of the solid is \( 1655\pi \) cubic units.

Would you like further clarification on any steps or more details?

---

Here are 5 related questions for practice:
1. How would you adjust the problem if the region were rotated about the y-axis?
2. What is the difference between the disk method and the shell method?
3. How would the volume change if the region were rotated about the line \( y = 1 \) instead of the x-axis?
4. What happens if we change the boundary from \( x = 5 \) to \( x = 3 \)?
5. How do you compute the surface area of the solid generated by the same rotation?

**Tip:** Always check if the rotation is about the x-axis or y-axis, as it influences the choice between using the disk or shell method.

Find the volume of the solid that results from rotating the region bounded by the graphs of y−6x−1=0, y=0, and x=5 about the x-axis. Write the exact answer. Do not round.

This problem demonstrates how to find the exact volume of a solid formed by rotating the region bounded by y−6x−1=0, y=0, and x=5 around the x-axis. Using the disk method, this solution involves definite integration and expands (6x + 1)^2 to compute the exact volume as 1655π cubic units.

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