# Centroid problem

1. Aug 31, 2009

### clairez93

1. The problem statement, all variables and given/known data

A blade on an industrial fan has the configuration of a semicircle attached to a trapezoid (see figure). Find the centroid of the blade.

2. Relevant equations

3. The attempt at a solution

My plan was to solve for the centroid of the trapezoid, then of the semi circle, then average the two. I got the wrong answer, however. Here is my work:

Trapezoid

$$M_{x} = 0$$ (since this is the axis of symmetry)
$$M_{y} = \int^{6}_{0} 2x(\frac{1}{6}x + 1)dx = 60$$
$$A_{trap} = \frac{(4+2)(6)}{2}$$

$$\overline{x} = \frac{60}{18} = \frac{10}{3}$$
$$\overline{y} = 0$$

Circle

$$M_{x} = 0$$
$$M_{y} = \int^{2}_{-2} \frac{\sqrt{4-y^{2}}+6}{2}(\sqrt{4-y^{2}}+6) dy = \frac{4(9\pi + 58)}{3}$$
$$A_{circle} = \frac{\pir^{2}}{2} = 2\pi$$

$$\overline{x} = \frac{\frac{4(9\pi + 58)}{3}}{2\pi} = \frac{2(9\pi + 58)}{3\pi}=$$
$$\overline{y} = 0$$

Averaging:

$$\frac{\frac{10}{3} + \frac{2(9\pi + 58)}{3\pi}}{2} = \frac{2(7\pi + 29)}{3\pi}$$
$$\overline{y} = 0$$

Book Answer: $$(\overline{x}, \overline{y}) = (\frac{2(9\pi + 49)}{3(\pi +9)}, 0)$$

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2. Sep 1, 2009

### HallsofIvy

The average of the centroids of two figures is NOT, in general, the centroid of their union. You need to use a "weighted" average, multiplying both numbers by the area of each figure, then dividing by the sum of the areas, not 2.