Centroid of an Area by Integration Determine the centroid of the triangular area in Fig. 7.4.

Question

Centroid of an Area by Integration

Determine the centroid of the triangular area in Fig. 7.4.

Strategy

We will determine the coordinates of the centroid by using an element of area dA in the form of a "strip" of width dx.

Accepted Answer

Let dA be the vertical strip in Fig. a. The height of the strip is {h/b)x, so dA = {h/b)x dx. To integrate over the entire area, we must integrate with respect to x from x = O to x = b. The x coordinate of the centroid is

[latex]\overline{x} = \frac{\int_{A}^{}{x dA} }{\int_{A}^{}{ dA}} = \frac{\int_{0}^{b}{x} \left(\frac{h}{b} x dx \right) }{\int_{0}^{b}{\frac{h}{b} x dx} } = \frac{\frac{h}{b} [\frac{x^{3} }{3} ]^{b}_{0} }{\frac{h}{b} [\frac{x^{2} }{2} ]^{b}_{0}} = \frac{2}{3} b.[/latex]

To determine [latex]\overline{y}[/latex] , we let y in Eq. (7.7) be the y coordinate of the midpoint of the strip (Fig. b):

[latex]\overline{y} = \frac{\int_{A}^{}{y d A} }{\int_{A}^{}{d A} } [/latex] (7.7)

[latex]\overline{y} = \frac{\int_{A}^{}{y dA} }{\int_{A}^{}{ dA}} = \frac{\int_{0}^{b}{\frac{1}{2} }\left(\frac{h}{b}x \right) \left(\frac{h}{b} x dx \right) }{\int_{0}^{b}{\frac{h}{b} x dx} } = \frac{\frac{1}{2} \left(\frac{h}{b} \right)^{2} [\frac{x^{3} }{3} ]^{b}_{0} }{\frac{h}{b} [\frac{x^{2} }{2} ]^{b}_{0}} = \frac{1}{3} h.[/latex]

The centroid is shown in Fig. c.

Discussion

You should always be alert for opportunities to check your results. In this example we should make sure that our integration procedure gives the correct result for the area of the triangle:

[latex]\int_{A}^{}{dA} = \int_{0}^{b}{\frac{h}{b} x dx} = \frac{h}{b} [\frac{x^{2} }{2} ]^{b}_{0} =\frac{1}{2} bh .[/latex]

Question 7.1: Centroid of an Area by Integration Determine the centroid of...

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