Find the coordinates of the centroid of the region bounded by 

y equals x squaredy=x2 

and 

y equals 4.y=4. 

The region is covered by a​ thin, flat plate.

To find the coordinates of the centroid of the region bounded by the curves \( y = x^2 \) and \( y = 4 \), we follow a similar process as before.

### 1. **Find the Points of Intersection:**

First, we determine where the curves intersect by setting \( y = x^2 \) equal to \( y = 4 \):

\[
x^2 = 4
\]

\[
x = \pm 2
\]

So, the curves intersect at the points \( (-2, 4) \) and \( (2, 4) \).

### 2. **Set Up the Integral for the Centroid:**

The centroid \((\bar{x}, \bar{y})\) of the region can be found using the formulas:

\[
\bar{x} = \frac{1}{A} \int_{a}^{b} x \left[ f(x) - g(x) \right] dx
\]

\[
\bar{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2} \left[ f(x)^2 - g(x)^2 \right] dx
\]

where:
- \( f(x) = 4 \) (the upper curve),
- \( g(x) = x^2 \) (the lower curve),
- \( a = -2 \), \( b = 2 \) (the limits of integration),
- \( A \) is the area of the region.

### 3. **Calculate the Area \( A \):**

The area \( A \) of the region is given by:

\[
A = \int_{-2}^{2} \left(4 - x^2\right) dx
\]

Calculating this integral:

\[
A = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2}
\]

\[
A = \left[ (4 \times 2) - \frac{2^3}{3} \right] - \left[ (4 \times -2) - \frac{(-2)^3}{3} \right]
\]

\[
A = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{-8}{3}\right)
\]

\[
A = \left(\frac{24}{3} - \frac{8}{3}\right) + \left(\frac{24}{3} - \frac{-8}{3}\right) = \frac{32}{3}
\]

### 4. **Calculate \( \bar{x} \):**

Now, we calculate \( \bar{x} \):

\[
\bar{x} = \frac{1}{A} \int_{-2}^{2} x \left(4 - x^2\right) dx
\]

Since \( \int_{-2}^{2} x \cdot (4 - x^2) dx \) is an odd function over a symmetric interval, the integral evaluates to 0:

\[
\bar{x} = 0
\]

### 5. **Calculate \( \bar{y} \):**

Now, we calculate \( \bar{y} \):

\[
\bar{y} = \frac{1}{A} \int_{-2}^{2} \frac{1}{2} \left[4^2 - (x^2)^2\right] dx
\]

\[
\bar{y} = \frac{3}{32} \cdot \frac{1}{2} \int_{-2}^{2} \left(16 - x^4\right) dx
\]

Breaking this into two integrals:

\[
\bar{y} = \frac{3}{32} \left[16x - \frac{x^5}{5}\right]_{-2}^{2}
\]

\[
\bar{y} = \frac{3}{32} \left(32 - \frac{32}{5}\right) = \frac{3}{32} \times \frac{128}{5} = \frac{12}{5}
\]

### **Final Answer:**
The centroid \((\bar{x}, \bar{y})\) of the region is:

\[
\left(0, \frac{12}{5}\right)
\]

Would you like any further details or have any questions? Here are five related questions you might find interesting:

1. How does the centroid change if the upper boundary is a different function?
2. What are some real-life applications of finding centroids in physics or engineering?
3. Can the method of finding centroids be extended to regions in three dimensions?
4. How does the symmetry of a region affect the location of its centroid?
5. How would you find the centroid if the region is bounded by a curve and a line?

**Tip:** When dealing with centroids, always check for symmetry in the region. Symmetry can simplify the computation of the centroid's coordinates.

Learn how to find the centroid of the region bounded by the curves y = x^2 and y = 4 using calculus and integration. This problem involves calculating the area and centroids under specified curves, suitable for college-level students studying calculus.

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