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.