Find the center of mass of the homogenous lamina in the first quadrant bounded by
y = 0, x = 0, f(y) = −y^2 + 9.

We are tasked with finding the center of mass (centroid) of a homogeneous lamina in the first quadrant, bounded by:

- \( y = 0 \)
- \( x = 0 \)
- \( f(y) = -y^2 + 9 \)

This problem involves calculating the center of mass of a region, given by the centroid coordinates \((\bar{x}, \bar{y})\), which are defined as:

\[
\bar{x} = \frac{1}{A} \int\int x \, dA
\]
\[
\bar{y} = \frac{1}{A} \int\int y \, dA
\]

Where \( A \) is the area of the lamina.

### Step 1: Defining the region

The curve \( f(y) = -y^2 + 9 \) describes a parabola that intersects the x-axis at \( y = 3 \). Since we are in the first quadrant, the region is bounded by \( y = 0 \), \( x = 0 \), and the curve.

### Step 2: Finding the area \( A \)

The area of the region can be found by integrating the length of the curve \( x = -y^2 + 9 \) over \( y \) from 0 to 3.

\[
A = \int_0^3 (-y^2 + 9) \, dy
\]

Let's compute this.

\[
A = \int_0^3 (-y^2 + 9) \, dy = \left[ -\frac{y^3}{3} + 9y \right]_0^3
\]
\[
A = \left( -\frac{27}{3} + 9(3) \right) - (0)
\]
\[
A = (-9 + 27) = 18
\]

So, the area \( A = 18 \).

### Step 3: Finding \( \bar{x} \)

The x-coordinate of the centroid is given by:

\[
\bar{x} = \frac{1}{A} \int_0^3 x(y) (-y^2 + 9) \, dy
\]

Since \( x(y) = -y^2 + 9 \), we are integrating \( (-y^2 + 9)^2 \) over \( y \).

\[
\bar{x} = \frac{1}{A} \int_0^3 (-y^2 + 9)^2 \, dy
\]

Let me compute this integral.The x-coordinate of the centroid is:

\[
\bar{x} = 7.2
\]

### Step 4: Finding \( \bar{y} \)

The y-coordinate of the centroid is given by:

\[
\bar{y} = \frac{1}{A} \int_0^3 y (-y^2 + 9) \, dy
\]

Let me compute this integral.The y-coordinate of the centroid is:

\[
\bar{y} = 1.125
\]

### Final Result:

The center of mass (centroid) of the lamina is located at:

\[
(\bar{x}, \bar{y}) = (7.2, 1.125)
\]

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. How would the center of mass change if the parabola was shifted?
2. How is the area calculated when the region is more complex?
3. What happens to the centroid if the lamina has varying density?
4. How can we calculate the moment of inertia for this lamina?
5. How does the center of mass shift if the lamina extends to other quadrants?

**Tip:** When finding centroids, the symmetry of the region can often simplify calculations.

Find the center of mass of the homogeneous lamina in the first quadrant bounded by y = 0, x = 0, f(y) = −y^2 + 9.

This guide explains how to find the center of mass (centroid) of a homogeneous lamina in the first quadrant, bounded by y = 0, x = 0, and f(y) = −y^2 + 9. The solution covers key calculus concepts, including centroid formulas and area under a curve, suitable for students in Grades 11-12.

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