The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by using shell method
y = −x2 + 17x − 72, y = 0;  about the y−axis

To find the volume of the solid formed by rotating the region bounded by \( y = -x^2 + 17x - 72 \) and \( y = 0 \) around the **y-axis** using the shell method, we follow these steps:

### 1. **Identify the curves and bounds**
We are given the curve:

\[
y = -x^2 + 17x - 72
\]
and the horizontal axis \( y = 0 \) as the bounds of the region.

We need to find where the curve intersects the x-axis, which will give us the limits for \( x \). This happens when \( y = 0 \).

\[
0 = -x^2 + 17x - 72
\]

Solve the quadratic equation for \( x \):

\[
x^2 - 17x + 72 = 0
\]

Using the quadratic formula:

\[
x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(1)(72)}}{2(1)}
\]
\[
x = \frac{17 \pm \sqrt{289 - 288}}{2}
\]
\[
x = \frac{17 \pm \sqrt{1}}{2}
\]
\[
x = \frac{17 \pm 1}{2}
\]

Thus, the solutions are:
\[
x = 9 \quad \text{and} \quad x = 8
\]

So, the region is bounded between \( x = 8 \) and \( x = 9 \).

### 2. **Set up the volume integral using the shell method**
The formula for the volume of a solid of revolution using the shell method is:

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

Here:
- The **radius** is the distance from the axis of rotation (the y-axis), which is simply \( x \).
- The **height** is the value of the function, which is \( y = -x^2 + 17x - 72 \).

So, the volume integral becomes:

\[
V = 2\pi \int_{8}^{9} x(-x^2 + 17x - 72) \, dx
\]

### 3. **Evaluate the integral**
First, expand the integrand:

\[
x(-x^2 + 17x - 72) = -x^3 + 17x^2 - 72x
\]

Now, integrate term by term:

\[
V = 2\pi \int_{8}^{9} (-x^3 + 17x^2 - 72x) \, dx
\]
\[
V = 2\pi \left[ -\frac{x^4}{4} + \frac{17x^3}{3} - 36x^2 \right]_{8}^{9}
\]

### 4. **Compute the definite integral**

Now, let's substitute the limits of integration:

- When \( x = 9 \):

\[
-\frac{9^4}{4} + \frac{17 \cdot 9^3}{3} - 36 \cdot 9^2
\]

\[
= -\frac{6561}{4} + \frac{17 \cdot 729}{3} - 36 \cdot 81
\]
\[
= -1640.25 + 4131 - 2916
\]
\[
= -1640.25 + 1215
\]
\[
= -425.25
\]

- When \( x = 8 \):

\[
-\frac{8^4}{4} + \frac{17 \cdot 8^3}{3} - 36 \cdot 8^2
\]

\[
= -\frac{4096}{4} + \frac{17 \cdot 512}{3} - 36 \cdot 64
\]
\[
= -1024 + \frac{8704}{3} - 2304
\]
\[
= -1024 + 2901.33 - 2304
\]
\[
= -426.67
\]

Now subtract the values:

\[
V = 2\pi \left[ -425.25 - (-426.67) \right]
\]
\[
V = 2\pi (1.42)
\]
\[
V = 2.84\pi
\]

Thus, the volume is approximately:

\[
V \approx 8.92 \, \text{cubic units}
\]

### Conclusion:
The volume of the solid formed by rotating the region around the y-axis is approximately \( 8.92 \pi \) cubic units.

---

Would you like to go over any part in more detail, or need help with a different concept?

Here are 5 related questions to expand your understanding:
1. How does the shell method differ from the disk/washer method in finding volumes?
2. How would the problem change if the region was rotated about a different axis, like the x-axis?
3. What is the significance of the limits of integration in determining the volume?
4. Can the volume of the same region be computed using the disk method? How?
5. How does the geometry of the region affect the complexity of the volume calculation?

**Tip:** When solving integrals in volume problems, break down the integrand carefully before attempting to integrate, to avoid mistakes!

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by using the shell method y = −x^2 + 17x − 72, y = 0; about the y−axis.

Learn how to find the volume of a solid of revolution using the shell method. The region bounded by the curve y = -x^2 + 17x - 72 and y = 0 is rotated about the y-axis, and the volume is calculated step by step. This problem involves concepts like the shell method, quadratic equations, and definite integrals, suitable for undergraduate calculus students.

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