Find the centroid of the region bounded by the given curves.
x = 10 − y2, x = 0

The problem asks for the centroid of the region bounded by the curves \( x = 10 - y^2 \) and \( x = 0 \).

### Step 1: Identify the region

The first equation, \( x = 10 - y^2 \), is a parabola that opens to the left, with its vertex at \( (10, 0) \). The second equation, \( x = 0 \), is the \( y \)-axis, meaning we are looking for the region between these two curves.

### Step 2: Find the bounds of integration

To find the bounds, we need to determine where the curves intersect, which occurs when \( x = 0 \):
\[
0 = 10 - y^2
\]
\[
y^2 = 10
\]
\[
y = \pm \sqrt{10}
\]

Thus, the region is bounded vertically between \( y = -\sqrt{10} \) and \( y = \sqrt{10} \).

### Step 3: Compute the area of the region

The formula for the area of the region is:
\[
A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy
\]
Here, \( x_{\text{right}} = 10 - y^2 \) and \( x_{\text{left}} = 0 \), so:
\[
A = \int_{-\sqrt{10}}^{\sqrt{10}} (10 - y^2) \, dy
\]

Now, calculate the integral:
\[
A = \int_{-\sqrt{10}}^{\sqrt{10}} 10 \, dy - \int_{-\sqrt{10}}^{\sqrt{10}} y^2 \, dy
\]
The first integral is:
\[
\int_{-\sqrt{10}}^{\sqrt{10}} 10 \, dy = 10 \cdot (2\sqrt{10}) = 20\sqrt{10}
\]
The second integral is:
\[
\int_{-\sqrt{10}}^{\sqrt{10}} y^2 \, dy = \frac{2}{3} \cdot (10^{3/2}) = \frac{20\sqrt{10}}{3}
\]

Thus, the total area is:
\[
A = 20\sqrt{10} - \frac{20\sqrt{10}}{3} = \frac{60\sqrt{10}}{3} - \frac{20\sqrt{10}}{3} = \frac{40\sqrt{10}}{3}
\]

### Step 4: Compute the centroid coordinates

The centroid \((\bar{x}, \bar{y})\) is given by the following formulas:

\[
\bar{x} = \frac{1}{A} \int_{y_1}^{y_2} x_{\text{right}} \cdot (x_{\text{right}} - x_{\text{left}}) \, dy
\]
\[
\bar{y} = \frac{1}{2A} \int_{y_1}^{y_2} y \cdot (x_{\text{right}} - x_{\text{left}}) \, dy
\]

#### Compute \( \bar{x} \):
\[
\bar{x} = \frac{1}{A} \int_{-\sqrt{10}}^{\sqrt{10}} (10 - y^2)(10 - y^2) \, dy
\]
Expanding:
\[
(10 - y^2)^2 = 100 - 20y^2 + y^4
\]
So:
\[
\bar{x} = \frac{1}{\frac{40\sqrt{10}}{3}} \int_{-\sqrt{10}}^{\sqrt{10}} (100 - 20y^2 + y^4) \, dy
\]
Each term integrates as follows:
\[
\int_{-\sqrt{10}}^{\sqrt{10}} 100 \, dy = 100 \cdot 2\sqrt{10} = 200\sqrt{10}
\]
\[
\int_{-\sqrt{10}}^{\sqrt{10}} 20y^2 \, dy = 20 \cdot \frac{20\sqrt{10}}{3} = \frac{400\sqrt{10}}{3}
\]
\[
\int_{-\sqrt{10}}^{\sqrt{10}} y^4 \, dy = \frac{2}{5} \cdot (10^{5/2}) = \frac{200\sqrt{10}}{5} = 40\sqrt{10}
\]

Thus, \( \bar{x} \) becomes:
\[
\bar{x} = \frac{3}{40\sqrt{10}} \left( 200\sqrt{10} - \frac{400\sqrt{10}}{3} + 40\sqrt{10} \right)
\]
Simplifying:
\[
\bar{x} = \frac{3}{40\sqrt{10}} \cdot \frac{200\sqrt{10} + 120\sqrt{10}}{3} = \frac{120}{40} = 3
\]

#### Compute \( \bar{y} \):
\[
\bar{y} = \frac{1}{2A} \int_{-\sqrt{10}}^{\sqrt{10}} y (10 - y^2) \, dy
\]
This is an odd function, so:
\[
\bar{y} = 0
\]

### Final answer:
The centroid of the region is \( (3, 0) \).

Do you want further details or have any questions? 

Here are some related problems to explore:
1. Find the centroid of a region bounded by \( y = x^2 \) and \( y = 0 \).
2. Find the area of the region bounded by \( y = \sqrt{x} \) and \( y = x - 2 \).
3. Compute the volume of a solid of revolution formed by rotating \( x = 10 - y^2 \) around the \( y \)-axis.
4. Determine the centroid of the region bounded by \( y = \sin x \) and the \( x \)-axis from \( x = 0 \) to \( x = \pi \).
5. Find the centroid of the region bounded by \( x^2 + y^2 = 9 \) and \( x = 0 \).

Tip: When finding centroids, check for symmetry—it can often simplify calculations.

Find the centroid of the region bounded by the curves x = 10 − y^2, x = 0.

Discover how to find the centroid of the region bounded by the curves x = 10 − y^2 and x = 0. This problem uses calculus concepts like definite integration and centroids to compute the area and centroid coordinates. Ideal for advanced high school and college calculus students.

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