Question

Set up and solve definite integrals to find each volume, surface area, or arc length that follows. Solve each volume problem both with disks/washers and with shells, if possible.

The centroid of the region between the graph of $f\left(x\right)=x^2$and the $x$-axis on $\left[0,2\right]$.

Short answer

$\overline{x}=\frac{3}{2},\overline{y}=\frac{6}{5}$.

Step-by-step solution

Step 1. Given Information.

The centroid of the region between the graph of $f\left(x\right)=x^2$and the $x$-axis on$[0,2].$

Step 2. Calculation of a mass and moments.

The total mass is :

$$\begin{gathered} m=\rho {\int }_a^{b}f\left(x\right)dx \\ m={\int }_0^2{x}^{2}dx \\ m={\left[\frac{x^3}{3}\right]}_0^2 \\ m=\frac{8}{3} \end{gathered}$$

The moments are :

$$\begin{gathered} M_x=\rho {\int }_a^b\frac{{\left[f\left(x\right)\right]}^2}{2}dx \\ {M}_x=\frac{1}{2}{\int }_0^2{x}^{4}dx \\ {M}_x={\left[\frac{x^5}{2\times 5}\right]}_0^2 \\ {M}_x=\frac{16}{5} \end{gathered}$$

$$\begin{gathered} M_y=\rho {\int }_a^{b}xf\left(x\right)dx \\ {M}_y={\int }_0^2{x}^{3}dx \\ {M}_y={\left[\frac{x^4}{4}\right]}_0^2 \\ {M}_y=\frac{16}{4}=4 \end{gathered}$$

Step 3. Centroid of a region.

We know that

$$\begin{gathered} \overline{x}=\frac{M_y}{m}=\frac{4}{{\displaystyle \frac{8}{3}}}=\frac{3}{2} \\ \overline{y}=\frac{M_x}{m}=\frac{{\displaystyle \frac{16}{5}}}{{\displaystyle \frac{8}{3}}}=\frac{16\times 3}{5\times 8}=\frac{6}{5}. \end{gathered}$$