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.