not x^2 let r be the region bounded above by y= 8-x  Let s be the solid of revolution by rtoating r about the line y=-3 find the volume

We are tasked with finding the volume of a solid formed by rotating the region \( r \), which is bounded above by \( y = 8 - x \), about the line \( y = -3 \).

### Step 1: Visualize the Problem
The region \( r \) is bounded by the line \( y = 8 - x \), which is a straight line with a slope of \(-1\) and a y-intercept of 8. Since there's no restriction on \( x \), we'll assume it’s bounded by the points where \( y = 8 - x \) intersects the x-axis (i.e., where \( y = 0 \)).

Solve for \( x \) when \( y = 0 \):
\[
8 - x = 0 \quad \Rightarrow \quad x = 8
\]
So, the region \( r \) is bounded from \( x = 0 \) to \( x = 8 \).

### Step 2: Set Up the Washer Method
Since we are rotating the region about the line \( y = -3 \), we'll use the washer method to calculate the volume. In this method, each cross-section perpendicular to the axis of rotation forms a washer (a disk with a hole).

The formula for the volume using the washer method is:
\[
V = \pi \int_a^b \left[ (R(x))^2 - (r(x))^2 \right] \, dx
\]
where:
- \( R(x) \) is the outer radius (the distance from the curve to the axis of rotation),
- \( r(x) \) is the inner radius (the distance from the x-axis to the axis of rotation),
- \( a \) and \( b \) are the bounds for \( x \), which in this case are 0 and 8.

### Step 3: Define the Radii
- The outer radius \( R(x) \) is the distance from the curve \( y = 8 - x \) to the line \( y = -3 \). The distance is given by:
  \[
  R(x) = (8 - x) - (-3) = 8 - x + 3 = 11 - x
  \]
- The inner radius \( r(x) \) is the distance from the x-axis (where \( y = 0 \)) to the line \( y = -3 \). This distance is:
  \[
  r(x) = 0 - (-3) = 3
  \]

### Step 4: Set Up the Integral
Now, substitute the expressions for \( R(x) \) and \( r(x) \) into the volume formula:
\[
V = \pi \int_0^8 \left[ (11 - x)^2 - 3^2 \right] \, dx
\]
Simplify:
\[
V = \pi \int_0^8 \left[ (11 - x)^2 - 9 \right] \, dx
\]

### Step 5: Expand and Solve the Integral
First, expand \( (11 - x)^2 \):
\[
(11 - x)^2 = 121 - 22x + x^2
\]
Now, substitute this back into the integral:
\[
V = \pi \int_0^8 \left[ 121 - 22x + x^2 - 9 \right] \, dx = \pi \int_0^8 \left[ 112 - 22x + x^2 \right] \, dx
\]
Now, integrate term by term:
\[
V = \pi \left[ 112x - 11x^2 + \frac{x^3}{3} \right]_0^8
\]

### Step 6: Evaluate the Integral
Substitute \( x = 8 \) and \( x = 0 \) into the expression:
\[
V = \pi \left[ 112(8) - 11(8)^2 + \frac{(8)^3}{3} \right]
\]
\[
V = \pi \left[ 896 - 11(64) + \frac{512}{3} \right]
\]
\[
V = \pi \left[ 896 - 704 + \frac{512}{3} \right]
\]
\[
V = \pi \left[ 192 + \frac{512}{3} \right]
\]
Now, find a common denominator:
\[
V = \pi \left[ \frac{576}{3} + \frac{512}{3} \right] = \pi \left[ \frac{1088}{3} \right]
\]
Thus, the volume is:
\[
V = \frac{1088\pi}{3} \ \text{cubic units}.
\]

### Conclusion:
The volume of the solid formed by rotating the region \( r \) about the line \( y = -3 \) is:
\[
V = \frac{1088\pi}{3} \ \text{cubic units}.
\]

Do you want more details on any part of this explanation? Here are 5 related questions you might want to explore:

1. How does the washer method differ from the disk method in finding volumes?
2. How do we determine the limits of integration in volume problems?
3. What is the general process for rotating a region around a horizontal or vertical line?
4. Can the volume be calculated using the shell method instead of the washer method?
5. How does changing the axis of rotation affect the volume?

**Tip**: Visualizing the region and the axis of rotation often helps in setting up the correct integral for volume problems.

not x^2 let r be the region bounded above by y= 8-x Let s be the solid of revolution by rotating r about the line y=-3 find the volume

Explore how to calculate the volume of a solid of revolution by rotating the region bounded by y = 8 - x about the line y = -3. This problem utilizes the washer method, with detailed steps and formulas. Ideal for students studying calculus in Grades 11-12 or college.

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