Find the centroid of the region bounded by the given curves.
y = x^4,    x = y^4

To find the centroid of the region bounded by the curves \( y = x^4 \) and \( x = y^4 \), we will proceed through several steps, including setting up integrals for the coordinates of the centroid. The centroid \((\bar{x}, \bar{y})\) is found using the formulas:

\[
\bar{x} = \frac{1}{A} \int_{x_1}^{x_2} x f(x) \, dx
\]
\[
\bar{y} = \frac{1}{A} \int_{y_1}^{y_2} y g(y) \, dy
\]
where \( A \) is the area of the region, and \( f(x) \) and \( g(y) \) are the upper and lower bounding curves in terms of \( x \) and \( y \), respectively.

### Step 1: Find the intersection points
We need to first find the points where the curves intersect, which will help us determine the limits of integration. The two equations are:

1. \( y = x^4 \)
2. \( x = y^4 \)

Substitute \( y = x^4 \) into \( x = y^4 \):

\[
x = (x^4)^4 = x^{16}
\]

This gives:

\[
x^{16} = x
\]

Factoring this equation:

\[
x(x^{15} - 1) = 0
\]

Thus, \( x = 0 \) or \( x = 1 \). The corresponding values of \( y \) are \( y = 0 \) when \( x = 0 \), and \( y = 1 \) when \( x = 1 \). So, the curves intersect at the points \( (0,0) \) and \( (1,1) \).

### Step 2: Set up the area integral
The curves divide the region between \( x = 0 \) and \( x = 1 \). For each \( x \), the upper curve is \( y = x^4 \) and the lower curve is \( y = x^{1/4} \). The area \( A \) of the region is:

\[
A = \int_0^1 (x^{1/4} - x^4) \, dx
\]

We can now compute this area integral.

### Step 3: Compute the area

\[
A = \int_0^1 x^{1/4} \, dx - \int_0^1 x^4 \, dx
\]

First integral:

\[
\int_0^1 x^{1/4} \, dx = \left[ \frac{x^{5/4}}{5/4} \right]_0^1 = \frac{4}{5}
\]

Second integral:

\[
\int_0^1 x^4 \, dx = \left[ \frac{x^5}{5} \right]_0^1 = \frac{1}{5}
\]

Thus, the total area is:

\[
A = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}
\]

### Step 4: Set up the centroid integrals
Now, we compute \( \bar{x} \) and \( \bar{y} \).

For \( \bar{x} \):

\[
\bar{x} = \frac{1}{A} \int_0^1 x (x^{1/4} - x^4) \, dx
\]

Breaking this into two integrals:

\[
\bar{x} = \frac{1}{\frac{3}{5}} \left( \int_0^1 x^{5/4} \, dx - \int_0^1 x^5 \, dx \right)
\]

First integral:

\[
\int_0^1 x^{5/4} \, dx = \left[ \frac{x^{9/4}}{9/4} \right]_0^1 = \frac{4}{9}
\]

Second integral:

\[
\int_0^1 x^5 \, dx = \left[ \frac{x^6}{6} \right]_0^1 = \frac{1}{6}
\]

Thus:

\[
\bar{x} = \frac{5}{3} \left( \frac{4}{9} - \frac{1}{6} \right) = \frac{5}{3} \times \frac{8 - 3}{18} = \frac{5}{3} \times \frac{5}{18} = \frac{25}{54}
\]

For \( \bar{y} \), due to symmetry (the equations are symmetric in \( x \) and \( y \)):

\[
\bar{y} = \bar{x} = \frac{25}{54}
\]

### Final Answer:
The centroid of the region is:

\[
\left( \frac{25}{54}, \frac{25}{54} \right)
\]

Do you have any questions, or would you like a detailed breakdown of any part?

Here are 5 related questions:
1. How is the formula for the centroid derived for more complex regions?
2. What happens if the curves involved do not intersect symmetrically?
3. Can centroids be found using polar coordinates for circular regions?
4. How would the calculation change if we rotated the region around an axis?
5. How can we calculate centroids for 3D volumes?

**Tip**: Symmetry often simplifies centroid calculations, as corresponding coordinates will have the same value.

Find the centroid of the region bounded by the given curves: y = x^4, x = y^4

Discover how to find the centroid of the region bounded by the curves y = x^4 and x = y^4 with a detailed solution. This problem involves calculus concepts like area integrals and symmetry, suitable for college-level calculus students.

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