problem 5.1 - locate the centroid of the plan area shown by integrati y y = 3x² 12 ft 2 ft x
problem 5.1 - locate the centroid of the plan area shown by integrati y y = 3x² 12 ft 2 ft x
Answer
Explanation:
Step1: Find the area (A)
The area (A) of the plane - area is given by the integral (A=\int_{a}^{b}y\mathrm{d}x). Here, (y = 3x^{2}), (a = 0), (b = 2). So (A=\int_{0}^{2}3x^{2}\mathrm{d}x). Using the power - rule for integration (\int x^{n}\mathrm{d}x=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (A=3\times\frac{x^{3}}{3}\big|{0}^{2}=x^{3}\big|{0}^{2}=2^{3}-0^{3}=8\mathrm{ft}^{2}).
Step2: Find the (x) - coordinate of the centroid (\bar{x})
The formula for (\bar{x}) is (\bar{x}=\frac{1}{A}\int_{a}^{b}x\cdot y\mathrm{d}x). Substitute (y = 3x^{2}) and (A = 8) into the formula: (\bar{x}=\frac{1}{8}\int_{0}^{2}x\cdot3x^{2}\mathrm{d}x=\frac{3}{8}\int_{0}^{2}x^{3}\mathrm{d}x). Using the power - rule for integration, (\frac{3}{8}\times\frac{x^{4}}{4}\big|_{0}^{2}=\frac{3}{32}(2^{4}-0^{4})=\frac{3}{32}\times16=\frac{3}{2}=1.5\mathrm{ft}).
Step3: Find the (y) - coordinate of the centroid (\bar{y})
The formula for (\bar{y}) is (\bar{y}=\frac{1}{A}\int_{a}^{b}\frac{y}{2}\cdot y\mathrm{d}x=\frac{1}{2A}\int_{a}^{b}y^{2}\mathrm{d}x). Substitute (y = 3x^{2}) and (A = 8) into the formula: (\bar{y}=\frac{1}{2\times8}\int_{0}^{2}(3x^{2})^{2}\mathrm{d}x=\frac{1}{16}\int_{0}^{2}9x^{4}\mathrm{d}x). Using the power - rule for integration, (\frac{9}{16}\times\frac{x^{5}}{5}\big|_{0}^{2}=\frac{9}{80}(2^{5}-0^{5})=\frac{9}{80}\times32=\frac{18}{5}=3.6\mathrm{ft}).
Answer:
The centroid of the plane area is ((\bar{x},\bar{y})=(1.5\mathrm{ft},3.6\mathrm{ft}))