Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the given curve about the y-axis. y = 9 - (x - 9)^2

The problem involves finding the volume of the solid obtained by rotating the region in the first quadrant bounded by the curve \( y = 9 - (x - 9)^2 \) about the y-axis. To solve this, we can use the method of cylindrical shells.

### Step-by-step Solution:

1. **Rewrite the equation**:
   The given equation is \( y = 9 - (x - 9)^2 \). This is a parabola shifted to the right along the x-axis, with its vertex at \( (9, 9) \).

2. **Find the limits of integration**:
   Since the region is bounded in the first quadrant, we need to find the \( x \)-intercepts of the curve. Set \( y = 0 \) and solve for \( x \):
   \[
   0 = 9 - (x - 9)^2
   \]
   \[
   (x - 9)^2 = 9
   \]
   Taking the square root of both sides:
   \[
   x - 9 = \pm 3
   \]
   Thus, \( x = 6 \) and \( x = 12 \).

   The region we are interested in is from \( x = 6 \) to \( x = 9 \) in the first quadrant.

3. **Use the cylindrical shells method**:
   For a rotation around the y-axis, the volume of a cylindrical shell is given by the formula:
   \[
   V = 2\pi \int_{x_1}^{x_2} x \cdot h(x) \, dx
   \]
   where \( h(x) = y \) is the height of the shell at any point \( x \).

   In our case, \( h(x) = 9 - (x - 9)^2 \), and the limits of integration are \( x_1 = 6 \) and \( x_2 = 12 \).

   So the volume is:
   \[
   V = 2\pi \int_{6}^{12} x \cdot \left(9 - (x - 9)^2 \right) dx
   \]

4. **Expand the integrand**:
   \[
   V = 2\pi \int_{6}^{12} x \cdot \left( 9 - (x^2 - 18x + 81) \right) dx
   \]
   Simplify the expression inside the integral:
   \[
   V = 2\pi \int_{6}^{12} x \cdot \left( 9 - x^2 + 18x - 81 \right) dx
   \]
   \[
   V = 2\pi \int_{6}^{12} x \cdot \left( -x^2 + 18x - 72 \right) dx
   \]
   Now distribute \( x \):
   \[
   V = 2\pi \int_{6}^{12} \left( -x^3 + 18x^2 - 72x \right) dx
   \]

5. **Integrate**:
   Now, integrate each term:
   \[
   \int -x^3 dx = -\frac{x^4}{4}, \quad \int 18x^2 dx = 6x^3, \quad \int -72x dx = -36x^2
   \]
   Applying these results, we get:
   \[
   V = 2\pi \left[ \left( -\frac{x^4}{4} + 6x^3 - 36x^2 \right) \Bigg|_6^{12} \right]
   \]

6. **Evaluate the definite integral**:
   Now, substitute \( x = 12 \) and \( x = 6 \) into the expression:

   For \( x = 12 \):
   \[
   \left( -\frac{12^4}{4} + 6(12)^3 - 36(12)^2 \right) = \left( -\frac{20736}{4} + 6(1728) - 36(144) \right)
   \]
   \[
   = -5184 + 10368 - 5184 = 0
   \]

   For \( x = 6 \):
   \[
   \left( -\frac{6^4}{4} + 6(6)^3 - 36(6)^2 \right) = \left( -\frac{1296}{4} + 6(216) - 36(36) \right)
   \]
   \[
   = -324 + 1296 - 1296 = -324
   \]

7. **Final calculation**:
   Now calculate the volume:
   \[
   V = 2\pi \left[ 0 - (-324) \right] = 2\pi \times 324 = 648\pi
   \]

Thus, the volume of the solid is \( \boxed{648\pi} \) cubic units.

Would you like further details on any step, or additional examples?

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Here are 5 related questions to explore further:
1. How do you find the volume of solids using the disk method?
2. What is the difference between the disk and shell methods?
3. How would the method change if rotating around the x-axis instead?
4. Can this problem be solved using polar coordinates?
5. How do you find the centroid of the region described?

**Tip:** The cylindrical shells method is typically easier to use when rotating around the y-axis, especially when the function is written as \( y = f(x) \).

Learn how to find the volume of the solid formed by rotating the curve y = 9 - (x - 9)^2 around the y-axis in the first quadrant. This problem covers key calculus concepts, such as the cylindrical shells method and definite integrals. Step-by-step solution provided for college-level students.

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