The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method.
y = −x2 + 11x − 30, y = 0;  about the y−axis
V =

To find the volume \( V \) of the solid generated by rotating the region bounded by the curve \( y = -x^2 + 11x - 30 \) and \( y = 0 \) around the y-axis, we will use the **method of cylindrical shells**. Here's the step-by-step process:

### Step 1: Identify the bounds of integration
First, we need to find the points where the curve \( y = -x^2 + 11x - 30 \) intersects the x-axis. This occurs where \( y = 0 \). Setting \( -x^2 + 11x - 30 = 0 \) gives the quadratic equation:

\[
-x^2 + 11x - 30 = 0
\]
Multiplying by \(-1\) to simplify:

\[
x^2 - 11x + 30 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -11 \), and \( c = 30 \). Plugging these values into the formula:

\[
x = \frac{11 \pm \sqrt{(-11)^2 - 4(1)(30)}}{2(1)}
\]
\[
x = \frac{11 \pm \sqrt{121 - 120}}{2}
\]
\[
x = \frac{11 \pm \sqrt{1}}{2}
\]
\[
x = \frac{11 \pm 1}{2}
\]

So, \( x = \frac{11 + 1}{2} = 6 \) or \( x = \frac{11 - 1}{2} = 5 \).

Thus, the region of integration is from \( x = 5 \) to \( x = 6 \).

### Step 2: Set up the formula for cylindrical shells
The formula for the volume using cylindrical shells when rotating about the y-axis is:

\[
V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx
\]

Here, \( f(x) = -x^2 + 11x - 30 \), and the bounds are \( x = 5 \) and \( x = 6 \).

### Step 3: Set up the integral
Substitute the function \( f(x) = -x^2 + 11x - 30 \) into the volume formula:

\[
V = 2\pi \int_{5}^{6} x \left(-x^2 + 11x - 30\right) \, dx
\]

### Step 4: Solve the integral
Now, expand the integrand:

\[
V = 2\pi \int_{5}^{6} \left(-x^3 + 11x^2 - 30x\right) \, dx
\]

Integrating each term:

\[
V = 2\pi \left[ -\frac{x^4}{4} + \frac{11x^3}{3} - 15x^2 \right]_{5}^{6}
\]

### Step 5: Evaluate the definite integral
Now, substitute \( x = 6 \) and \( x = 5 \) into the antiderivative:

For \( x = 6 \):

\[
-\frac{6^4}{4} + \frac{11(6^3)}{3} - 15(6^2) = -\frac{1296}{4} + \frac{11(216)}{3} - 15(36)
\]
\[
= -324 + 792 - 540 = -324 + 252 = -72
\]

For \( x = 5 \):

\[
-\frac{5^4}{4} + \frac{11(5^3)}{3} - 15(5^2) = -\frac{625}{4} + \frac{11(125)}{3} - 15(25)
\]
\[
= -156.25 + \frac{1375}{3} - 375 = -156.25 + 458.33 - 375
\]
\[
= -156.25 + 83.33 = -72.92
\]

Now, subtract the values:

\[
V = 2\pi \left(-72 - (-72.92)\right) = 2\pi(0.92) = 1.84\pi
\]

### Final Answer:
The volume \( V \) of the solid is approximately:

\[
V \approx 1.84\pi \, \text{cubic units}
\]

Would you like more details on any part of the process, or have any other questions?

### Related Questions:
1. How do you use the method of washers for similar problems?
2. How does the volume change if rotated around the x-axis instead?
3. What happens if the curve is rotated around the line \( x = 3 \)?
4. Can this problem be solved using the disk method, and how?
5. What would the integral look like if the bounds were different?

### Tip:
The method of cylindrical shells is especially useful when rotating around the y-axis and dealing with functions of \( x \).

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method.
 y = −x^2 + 11x − 30, y = 0;  about the y−axis
 V =

Discover how to calculate the volume of a solid formed by rotating the curve y = −x^2 + 11x − 30 and y = 0 around the y-axis. This problem uses the cylindrical shells method and requires solving a quadratic equation for the bounds of integration. Suitable for college-level calculus or AP calculus students.

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