Determine the centroid the region in the plane which lies above the x-axis and b

Question

Determine the centroid the region in the plane which lies above the x-axis and below the semi-circle y =

r2?x2????????

Explanation / Answer

We know that the mass (m) of a planar lamina such that f(x) = g(x) on [a, b] with uniform density (p) is: m = p ? [f(x) - g(x)] dx (from x=a to b). Then, the moments about the x-axis (Mx) and y-axis (My) is: Mx = p ? [f(x) + g(x)]/2 * [f(x) - g(x)] dx (from x=a to b) My = p ? x[f(x) - g(x)] dx (from x=a to b) Then, the centroid is given by: ? ? (x, y) Where: ? x. . . .= (My)/m ? y. . . .= (Mx)/m We see that the points of intersection of y = x + 2 and y = x^2 occurs when the y-values are equal; or: x^2 = x + 2 ==> x^2 - x - 2 = 0 ==> (x - 2)(x + 1) = 0 ==> x = -1, 2. So we are considering the interval of [-1, 2]. However, note that x + 2 = x^2 for all x E [-1, 2]. So f(x) = x + 2 and g(x) = x^2. So: Mx = p ? [f(x) + g(x)]/2 * [f(x) - g(x)] dx (from x=a to b) . . ..= p ? (x^2 + x + 2)/2 * (x + 2 - x^2) dx (from x=-1 to 2) . . ..= p ? [(-1/2)x^4 + (1/2)x^2 + 2x + 2] dx (from x=-1 to 2) . . ..= p * [(-1/10)x^5 + (1/6)x^3 + x^2 + 2x] (evaluated from x=-1 to 2) . . ..= p * {[(-1/10)(32) + (1/6)(8) + 4 + 4] - [(-1/10)(-1) + (1/6)(-1) + 1 - 2]} . . ..= p * [92/15 - (-16/15)] . . ..= (36/5)p. My = p ? x[f(x) - g(x)] dx (from x=a to b) . . ..= p ? x(x + 2 - x^2) dx (from x=-1 to 2) . . ..= p ? (-x^3 + x^2 + 2x) dx (from x=-1 to 2) . . ..= p * [(-1/4)x^4 + (1/3)x^3 + x^2] (evaluated from x=-1 to 2) . . ..= p * {[(-1/4)(16) + (1/3)(8) + 4] - [(-1/4)(1) + (1/3)(-1) + 1]} . . ..= (9/4)p. m = p ? [f(x) - g(x)] dx (from x=a to b) . . = p ? x + 2 - x^2 dx (from x=-1 to 2) . . = p * [(-1/3)x^3 + (1/2)x^2 + 2x] (evaluated from x=-1 to 2) . . = p * {[(-1/3)(8) + (1/2)(4) + 4] - [(-1/3)(-1) + (1/2)(1) - 2]} . . = (9/2)p. Thus, the coordinates of the centroid are: [(My)/m, (Mx)/m] = [(9/4)/(9/2), (36/5)/(9/2)] = (1/2, 8/5).

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