Find the center of mass of the homogenous lamina in the first quadrant bounded by 
y = 0,
 
x = 0,
 
f(y) = −y2 + 16.

To find the center of mass of the homogeneous lamina, we need to determine the coordinates \((\bar{x}, \bar{y})\), which represent the center of mass in the \(x\)- and \(y\)-directions, respectively. The lamina is bounded by the curve \(f(y) = -y^2 + 16\), the \(x\)-axis, and the \(y\)-axis in the first quadrant. 

### 1. Setting Up the Region:
The boundary curve is given by \(x = f(y) = -y^2 + 16\). Since the lamina is in the first quadrant, we have:
- \(x = 0\) (the \(y\)-axis),
- \(y = 0\) (the \(x\)-axis).

We find the range of \(y\) where the curve is in the first quadrant:
\[
x = -y^2 + 16 \geq 0 \quad \Rightarrow \quad y^2 \leq 16 \quad \Rightarrow \quad -4 \leq y \leq 4.
\]
Since we are only in the first quadrant, \(y \in [0, 4]\).

### 2. Formulas for the Center of Mass:
The center of mass is given by the coordinates:
\[
\bar{x} = \frac{1}{A} \int_{\text{Region}} x \, dA, \quad \bar{y} = \frac{1}{A} \int_{\text{Region}} y \, dA,
\]
where \(A\) is the area of the region.

- **Area** (\(A\)): 
The area is given by:
\[
A = \int_0^4 \left( -y^2 + 16 \right) \, dy.
\]
Let's compute this area:
\[
A = \int_0^4 ( -y^2 + 16 ) \, dy = \left[ -\frac{y^3}{3} + 16y \right]_0^4.
\]
Substitute \(y = 4\) and \(y = 0\):
\[
A = \left( -\frac{4^3}{3} + 16 \cdot 4 \right) - 0 = \left( -\frac{64}{3} + 64 \right) = \frac{192}{3} - \frac{64}{3} = \frac{128}{3}.
\]

### 3. Computing \(\bar{x}\):
The formula for \(\bar{x}\) is:
\[
\bar{x} = \frac{1}{A} \int_0^4 x(y) \cdot x' \, dy = \frac{1}{A} \int_0^4 (-y^2 + 16)(-y^2 ) you'd_length And this halfant

Find the center of mass of the homogenous lamina in the first quadrant bounded by y = 0, x = 0, f(y) = −y^2 + 16.

Explore how to calculate the center of mass for a homogeneous lamina in the first quadrant, bounded by the curve f(y) = -y^2 + 16. This problem involves applying concepts from calculus, including definite integrals and the center of mass formula. Suitable for undergraduate calculus students.

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