Q. 15. - Calculus

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 graphs of $f\left(x\right) = \sqrt{x}$ and $g\left(x\right) = 4 - x$ on $[0, 4]$ .

Step-by-Step Solution

Answer

$\overline{x}=-\frac{4}{5} , \overline{y }=-1$.

Step 1. Given Information.

The centroid of the region between the graphs of $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=4-x$ on $\left[0,4\right]$ .

Step 2. Calculation of mass and moments.

The total mass is :

$$\begin{gathered} m=\rho {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]dx \\ m={\int }_0^4\left[\sqrt{x}-\left(4-x\right)\right]dx \\ m={\int }_0^4\left[\sqrt{x}-4+x\right]dx \\ m={\left[\frac{2x^{{\displaystyle \frac{3}{2}}}}{3}-4x+\frac{x^2}{2}\right]}_0^4 \\ m=-\frac{8}{3} \end{gathered}$$

The moments are :
$$\begin{gathered} M_x=\rho {\int }_a^b\frac{{\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2}{2}dx \\ {M}_x=\frac{1}{2}{\int }_0^4{\left[\sqrt{x}\right]}^2-{\left[4-x\right]}^{2}dx \\ {M}_x=\left[\frac{1}{2}\right]{\int }_0^{4}x-\left[16+x^2-8x\right]dx \\ {M}_x=\frac{1}{2}{\int }_0^4\left[x-16-x^2+9x\right] dx \\ {M}_x=\frac{1}{2}{\int }_0^4\left[-16-x^2+10x\right] dx \\ {M}_x=\frac{1}{2}{\left[-16x-\frac{x^3}{3}+\frac{10x^2}{2}\right]}_0^4 \\ {M}_x=\frac{8}{3} \end{gathered}$$

$$\begin{gathered} M_y=\rho {\int }_a^{b}x\left[f\left(x\right) - g\left(x\right)\right]dx \\ {M}_y={\int }_0^{4}x\left[\sqrt{x}-\left(4-x\right)\right]dx \\ {M}_y={\int }_0^4\left[x^{\frac{3}{2}}-4x+x^2\right]dx \\ {M}_y={\left[\frac{2x^{\frac{5}{2}}}{5}-\frac{4x^2}{2}+\frac{x^3}{3}\right]}_0^4 \\ {M}_y=\frac{32}{15}. \end{gathered}$$

Step 3. Centroid of a given region.

The centroid of a given region are :

$$\begin{gathered} \overline{x} = \frac{M_y}{m}=\frac{{\displaystyle \frac{32}{15}}}{{\displaystyle -\frac{8}{3}}}=-\frac{32\times 3}{8\times 15}=-\frac{4}{5}. \\ \overline{y }=\frac{M_x}{m}=-\frac{{\displaystyle \frac{8}{3}}}{{\displaystyle \frac{8}{3}}}=-1. \end{gathered}$$