Find the centroid of the region bounded by the given curves.

Question

Quizplus · Accepted Answer

The region is bounded by the curves $y=x^2$ and $y=3-x^2$. We can find the $x$-values where the two curves intersect by setting $x^2=3-x^2$ and solving for $x$, which gives us $x=\pm 1$.

To find the $y$-coordinate of the centroid, we can use the formula:

$$\bar{y} = \frac{1}{A} \int_{a}^{b} \int_{f(x)}^{g(x)} y\,dy\,dx$$

where $a=-1$, $b=1$, $f(x)=x^2$, $g(x)=3-x^2$, and $A$ is the area of the region.

To find $A$, we can integrate from $-1$ to $1$ using the formula:

$$A=\int_{a}^{b} [g(x)-f(x)]\,dx$$

Putting everything together, we have:

$$A=\int_{-1}^{1} [(3-x^2)-x^2]\,dx = \int_{-1}^{1} (3-2x^2)\,dx = 4$$

$$\bar{y} = \frac{1}{4} \int_{-1}^{1} \int_{x^2}^{3-x^2} y\,dy\,dx = \frac{1}{4} \int_{-1}^{1} \left[\frac{y^2}{2}\right]_{x^2}^{3-x^2} dx$$

$$\bar{y} = \frac{1}{4} \int_{-1}^{1} \frac{(3-x^2)^2}{2}-\frac{x^4}{2}\, dx = \frac{1}{4} \int_{-1}^{1} \frac{9-6x^2+x^4}{2}-\frac{x^4}{2}\, dx$$

$$\bar{y} = \frac{1}{4} \int_{-1}^{1} \frac{9-7x^2}{2}\, dx = 0$$

Therefore, the centroid is located on the $x$-axis, and we can confirm that it is at $(0,0)$ by symmetry.

Find the Centroid of the Region Bounded by the Given