problem 5.1 - locate the centroid of the plan area shown by integration.

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). Using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (A=\left[x^{3}\right]_{0}^{2}=8\mathrm{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). Using the power - rule, (M_y=\left[\frac{3x^{4}}{4}\right]_{0}^{2}=12\mathrm{ft}^{3}). And (\bar{x}=\frac{M_y}{A}=\frac{12}{8}=\frac{3}{2}\mathrm{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). Using the power - rule, (M_x=\frac{9}{2}\left[\frac{x^{5}}{5}\right]_{0}^{2}=\frac{9}{2}\times\frac{32}{5}=\frac{144}{5}\mathrm{ft}^{3}). And (\bar{y}=\frac{M_x}{A}=\frac{\frac{144}{5}}{8}=\frac{18}{5} = 3.6\mathrm{ft}).

Answer:

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