A hemispherical bowl of radius 9 inches is filled to a depth of h​ inches, where h is between 0 and 9. 
Find the volume of water in the bowl as a function of h.

To find the volume of water in a hemispherical bowl of radius \( R = 9 \) inches, filled to a depth \( h \), we can approach this using calculus and the method of disk integration. Here's the step-by-step process:

### 1. Cross-sectional setup

A vertical cross-section of the hemisphere is a semicircle with radius \( R = 9 \). The equation of this semicircle is given by:
\[
x^2 + y^2 = R^2 \quad \text{or} \quad y = \sqrt{R^2 - x^2}
\]
where \( y \) is the height at a point \( x \) along the radius.

The bowl is filled to a height \( h \). So, the goal is to calculate the volume of the water in the bowl from \( y = 0 \) to \( y = h \).

### 2. Disk method setup

We'll use the method of disks. The radius of each disk at height \( y \) is determined by the cross-sectional distance from the center, which is \( x = \sqrt{R^2 - y^2} \). The volume of a thin disk at height \( y \) with thickness \( dy \) is:
\[
dV = \pi \left( \text{radius of the disk} \right)^2 \, dy = \pi \left( \sqrt{R^2 - y^2} \right)^2 \, dy = \pi (R^2 - y^2) \, dy
\]

### 3. Setting up the integral

To find the total volume of water up to a height \( h \), integrate the volume of these disks from \( y = 0 \) to \( y = h \):
\[
V(h) = \int_0^h \pi (R^2 - y^2) \, dy
\]

### 4. Solving the integral

We now compute the integral:
\[
V(h) = \pi \int_0^h (R^2 - y^2) \, dy = \pi \left[ R^2y - \frac{y^3}{3} \right]_0^h
\]
Substituting the limits:
\[
V(h) = \pi \left[ R^2h - \frac{h^3}{3} \right]
\]

### 5. Substituting \( R = 9 \)

Substitute \( R = 9 \) into the equation:
\[
V(h) = \pi \left[ 9^2h - \frac{h^3}{3} \right]
\]
\[
V(h) = \pi \left[ 81h - \frac{h^3}{3} \right]
\]

Thus, the volume of water in the bowl as a function of \( h \) is:
\[
V(h) = \pi \left( 81h - \frac{h^3}{3} \right) \quad \text{cubic inches}
\]

This is the required formula for the volume of water in the hemispherical bowl as a function of \( h \), where \( h \) is between 0 and 9 inches.

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Do you want any further clarification or details on the process?

### Related Questions:
1. How would the formula change if the bowl had a different radius \( R \)?
2. Can you calculate the volume of water when the bowl is filled to \( h = 6 \) inches?
3. What happens to the volume as \( h \to R \)?
4. How can this problem be generalized for a bowl with a different shape (e.g., parabolic)?
5. How can you calculate the surface area of the water at any given depth \( h \)?

### Tip:
Using symmetry and geometric properties of common shapes can often simplify integration in volume problems.

A hemispherical bowl of radius 9 inches is filled to a depth of h inches, where h is between 0 and 9. Find the volume of water in the bowl as a function of h.

Discover how to calculate the volume of water in a hemispherical bowl of radius 9 inches filled to a depth of h inches using the disk integration method. This solution involves calculus, disk method formulas, and integral calculus, suitable for Grades 11-12 and college-level students.

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