problem 5.6 - locate the centroid of the plan area shown by integration.

Explanation:

Step1: Find the area A

The area of a triangle is $A=\int_{0}^{b}y\mathrm{d}x$. Given $y = \frac{h}{b}x$, then $A=\int_{0}^{b}\frac{h}{b}x\mathrm{d}x$. Using the power - rule of integration $\int x^n\mathrm{d}x=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $A=\frac{h}{b}\int_{0}^{b}x\mathrm{d}x=\frac{h}{b}\left[\frac{x^{2}}{2}\right]_{0}^{b}=\frac{1}{2}bh$.

Step2: Find the x - coordinate of the centroid $\bar{x}$

The formula for $\bar{x}$ is $\bar{x}=\frac{1}{A}\int_{0}^{b}x\cdot y\mathrm{d}x$. Substitute $y=\frac{h}{b}x$ and $A = \frac{1}{2}bh$ into the formula: $\bar{x}=\frac{2}{bh}\int_{0}^{b}x\cdot\frac{h}{b}x\mathrm{d}x=\frac{2h}{b^{2}h}\int_{0}^{b}x^{2}\mathrm{d}x=\frac{2}{b^{2}}\left[\frac{x^{3}}{3}\right]_{0}^{b}=\frac{2}{3}b$.

Step3: Find the y - coordinate of the centroid $\bar{y}$

The formula for $\bar{y}$ is $\bar{y}=\frac{1}{A}\int_{0}^{b}\frac{y}{2}\cdot y\mathrm{d}x$. Substitute $y = \frac{h}{b}x$ and $A=\frac{1}{2}bh$ into the formula: $\bar{y}=\frac{2}{bh}\int_{0}^{b}\frac{1}{2}\left(\frac{h}{b}x\right)^{2}\mathrm{d}x=\frac{1}{bh}\cdot\frac{h^{2}}{b^{2}}\int_{0}^{b}x^{2}\mathrm{d}x=\frac{h}{b^{3}}\left[\frac{x^{3}}{3}\right]_{0}^{b}=\frac{1}{3}h$.

Answer:

The centroid is $(\bar{x},\bar{y})=\left(\frac{2}{3}b,\frac{1}{3}h\right)$