problem 5.1 - locate the centroid of the plan area shown by integration. y y = 3x² 12 ft x 2 ft

Explanation:

Step1: Find the area $A$

The area of a differential element $dA = ydx=3x^{2}dx$. Integrating from $x = 0$ to $x = 2$ to find the total area: $$A=\int_{0}^{2}3x^{2}dx=\left[x^{3}\right]_{0}^{2}=8\text{ ft}^2$$

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

The moment about the $y$-axis $M_y=\int_{A}x dA$. Since $dA = 3x^{2}dx$, then $M_y=\int_{0}^{2}x\cdot3x^{2}dx=\int_{0}^{2}3x^{3}dx=\left[\frac{3}{4}x^{4}\right]_{0}^{2}=12$. And $\bar{x}=\frac{M_y}{A}=\frac{12}{8}=\frac{3}{2}\text{ ft}$

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

The moment about the $x$-axis $M_x=\int_{A}\frac{y}{2}dA$. Since $y = 3x^{2}$ and $dA=3x^{2}dx$, then $M_x=\frac{1}{2}\int_{0}^{2}(3x^{2})\cdot3x^{2}dx=\frac{9}{2}\int_{0}^{2}x^{4}dx=\frac{9}{2}\left[\frac{1}{5}x^{5}\right]_{0}^{2}=\frac{144}{5}$. And $\bar{y}=\frac{M_x}{A}=\frac{\frac{144}{5}}{8}=\frac{18}{5} = 3.6\text{ ft}$

Answer:

The centroid is at $(\bar{x},\bar{y})=\left(\frac{3}{2}\text{ ft},3.6\text{ ft}\right)$