I have this eqn y = x^2 -9, now it is known that the area bounded by x axis, y= h and the curve, when rotate around y axis 180 deg generates a volume of 40 pi , determine h, which is between 0 to -9

To solve for \( h \), we need to calculate the volume of the solid generated by rotating the area bounded by the curve \( y = x^2 - 9 \), the x-axis, and \( y = h \) around the y-axis. We are given that this volume is \( 40\pi \).

### Step 1: Express \( x^2 \) in terms of \( y \)
Given \( y = x^2 - 9 \), we can rewrite this as:
\[
x^2 = y + 9
\]

### Step 2: Set up the volume integral
The volume \( V \) of the solid generated by rotating the curve around the y-axis is given by the formula:
\[
V = \pi \int_{y_1}^{y_2} R(y)^2 \, dy
\]
Where \( R(y) \) is the radius of rotation at a given \( y \), which corresponds to \( x \) in this case. So, \( R(y) = \sqrt{y + 9} \).

Here, \( y_1 = h \) and \( y_2 = 0 \).

The volume formula becomes:
\[
V = \pi \int_{h}^{0} (\sqrt{y + 9})^2 \, dy
\]

### Step 3: Simplify the integral
Simplifying the expression inside the integral:
\[
V = \pi \int_{h}^{0} (y + 9) \, dy
\]

### Step 4: Compute the integral
Now, let's integrate:
\[
V = \pi \left[ \frac{y^2}{2} + 9y \right]_{h}^{0}
\]
Evaluating this at the bounds:
\[
V = \pi \left[ \left( \frac{0^2}{2} + 9(0) \right) - \left( \frac{h^2}{2} + 9h \right) \right]
\]
Simplifying further:
\[
V = -\pi \left( \frac{h^2}{2} + 9h \right)
\]
Given that \( V = 40\pi \), we equate and solve for \( h \):
\[
-\pi \left( \frac{h^2}{2} + 9h \right) = 40\pi
\]
\[
-\left( \frac{h^2}{2} + 9h \right) = 40
\]
\[
\frac{h^2}{2} + 9h + 40 = 0
\]
Multiply through by 2 to get rid of the fraction:
\[
h^2 + 18h + 80 = 0
\]

### Step 5: Solve the quadratic equation
To solve the quadratic equation:
\[
h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 18 \), and \( c = 80 \). Plugging in these values:
\[
h = \frac{-18 \pm \sqrt{18^2 - 4(1)(80)}}{2(1)}
\]
\[
h = \frac{-18 \pm \sqrt{324 - 320}}{2}
\]
\[
h = \frac{-18 \pm \sqrt{4}}{2}
\]
\[
h = \frac{-18 \pm 2}{2}
\]
This gives us two possible solutions:
\[
h = \frac{-16}{2} = -8 \quad \text{or} \quad h = \frac{-20}{2} = -10
\]
But since \( h \) must be between 0 and -9, the correct solution is \( h = -8 \).

### Final Answer:
\( h = -8 \).

Would you like to dive deeper into any part of the solution or have any further questions?

Here are 5 related questions:
1. How would the problem change if the rotation was around the x-axis?
2. What is the significance of the volume formula used here?
3. How would the volume be different if \( h \) was closer to 0?
4. What is the geometrical interpretation of rotating a curve around an axis?
5. Can this problem be solved using cylindrical shells?

**Tip:** When solving for variables in a quadratic equation, always check which solution fits the context of the problem, as some solutions might be extraneous.

Learn how to find the value of h in the equation y = x^2 - 9 that generates a volume of 40π when rotated around the y-axis. This problem involves the use of calculus concepts such as the disc method and solving quadratic equations. Suitable for students in Grades 11-12.

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