# centroids - Centroids Consider a region in the plane of...

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Unformatted text preview: Centroids Consider a region in the plane of area A. We can think of the region as a thin plate with uniform thickness and density. The centroid of the region has coordinates (x, y). It can be found using x= 1 A xe dA and y = A 1 A ye dA A where (xe , ye ) is the coordinates of the centroid of the diﬀerential element of area dA. Example 1. Find the centroid of the region bounded by the curve y = x3 and the lines y = 0 and x = 2. Solution. We will use diﬀerential elements consisting of rectangular vertical slices of height y and width dx. This means that variable x will be the variable of integration. y y = x3 8 (xe , ye ) y 0 dx x 2 x We ﬁrst ﬁnd the area A. 2 A= dA = A 2 x3 dx = y dx = 0 0 2 x4 4 = 0 dA 24 =4 4 Now, observe that the centroid of the diﬀerential element has coordinates (xe , ye ) where y xe = x and ye = . 2 Therefore, 2 2 2 2 x5 25 32 xe dA = x y dx = x(x3 ) dx = x4 dx = = = 5 0 5 5 A 0 0 0 dA and 2 ye dA = A 0 y y dx = 2 2 0 y2 dx = 2 2 0 (x3 )2 dx = 2 2 0 x6 x7 dx = 2 14 dA We can now ﬁnd the coordinates of the centroid. 1 32/5 8 x= xe dA = = A A 4 5 and y= We conclude that the centroid is located at the following point. (x, y) = Gilles Cazelais. Typeset with L T X on May 22, 2008. A E 8 16 , 5 7 1 A ye dA = A 2 = 0 27 64 = . 14 7 64/7 16 = 4 7 Example 2. Find the centroid of the region bounded by the curve y = √ x and the lines x = 0 and y = 3. Solution. We will use diﬀerential elements consisting of rectangular horizontal slices of height x and width dy. This means that variable y will be the variable of integration. y (xe , ye ) 3 y= dy x y 0 y= √ x implies that x = y 2 . Let’s ﬁnd the area A. 3 dA = A x 9 x Observe that A= √ 3 y 2 dy = x dy = 0 0 3 y3 3 = 0 dA 33 =9 3 Now, observe that the centroid of the diﬀerential element has coordinates (xe , ye ) where x 2 xe = and ye = y. Therefore, 3 xe dA = A 0 x x dy = 2 3 0 x2 dy = 2 3 0 (y 2 )2 dy = 2 3 0 y4 y5 dy = 2 10 dA and 3 ye dA = A 3 0 3 y(y 2 ) dy = y x dy = 0 y 3 dy = 0 y4 4 dA 3 = 0 3 = 0 35 243 = 10 10 34 81 = . 4 4 We can now ﬁnd the coordinates of the centroid. x= 1 A xe dA = A 243/10 27 = 9 10 and y= We conclude that the centroid is located at the following point. 27 9 , 10 4 (x, y) = 2 1 A ye dA = A 81/4 9 = 9 4 Let’s now consider the area between two curves. First consider the area bounded by the curves y = f (x) and y = g(x) over a ≤ x ≤ b. y (xe , ye ) y = f (x) x y = g(x) a x b dx In this case, we have xe = x and ye = f (x) + g(x) . 2 Therefore, b A= (f (x) − g(x)) dx dA = a A dA b A b x (f (x) − g(x)) dx xe dA = and f (x) + g(x) 2 ye dA = a A a (f (x) − g(x)) dx . dA dA Now consider the area bounded by the curves x = f (y) and x = g(y) over c ≤ y ≤ d. y x = g(y) x = f (y) d (xe , ye ) dy y x c In this case, we have xe = f (y) + g(y) 2 and ye = y. Therefore, d A= (f (y) − g(y)) dy dA = A c dA d xe dA = A c f (y) + g(y) 2 d (f (y) − g(y)) dy and A dA y (f (y) − g(y)) dy ye dA = c dA 3 Example 3. Find the centroid of the region between the curves y = 2x and y = x2 over the interval 0 ≤ x ≤ 2. Solution. We will use diﬀerential elements consisting of rectangular vertical slices of width dx. This means that variable x will be the variable of integration. y 4 y = 2x (xe , ye ) x y = x2 0 dx x 2 Let’s ﬁnd the area A. 2 A= (2x − x2 ) dx = dA = A x2 − 0 2 x3 3 =4− 0 8 4 = 3 3 dA Now, observe that 2x + x2 . 2 xe = x and ye = Therefore, 2 A 2 x (2x − x2 ) dx = xe dA = (2x2 − x3 ) dx = 0 0 2x3 x4 − 3 4 2 = 0 16 4 −4= 3 3 dA and 2 ye dA = A 0 2x + x2 2 2 (2x − x2 ) dx = 0 4x2 − x4 dx = 2 2x3 x5 − 3 10 2 = 0 16 16 32 − = . 3 5 15 dA We can now ﬁnd the coordinates of the centroid. x= 1 A xe dA = A 4/3 =1 4/3 and y= We conclude that the centroid is located at the following point. (x, y) = 4 1, 8 5 1 A ye dA = A 32/15 8 = 4/3 5 Example 4. Find the centroid of the region bounded by the curves y = 3 − x, y = √ x − 1, and y = 0. Solution. We will use diﬀerential elements consisting of horizontal vertical slices of width dy. This means that variable y will be the variable of integration. y y =3−x 2 y= (xe , ye ) √ x−1 1 dy y 0 1 x 3 √ Let’s ﬁnd the point of intersection of y = 3 − x and y = x − 1. √ 3 − x = x − 1 =⇒ 9 − 6x + x2 = x − 1 =⇒ x2 − 7x + 10 = 0 =⇒ (x − 2)(x − 5) = 0 We conclude that x = 2 is the solution. Then, the point of intersection is (2, 1). Now, observe that √ y = x − 1 =⇒ x = y 2 + 1 and y = 3 − x =⇒ x = 3 − y. We can describe the region as the region between the curves x = y 2 + 1 and x = 3 − y over 0 ≤ y ≤ 1. Let’s ﬁnd the area A. 1 A= 1 (3 − y) − (y 2 + 1) dy = dA = A (2 − y − y 2 ) dy = 0 2y − 0 1 y3 y2 − 2 3 = 0 7 6 dA Now, observe that xe = (3 − y) + (y 2 + 1) 2 and ye = y. Therefore, 1 xe dA = A 0 1 (3 − y) + (y 2 + 1) 2 (3 − y)2 − (y 2 + 1)2 dy 2 (3 − y) − (y 2 + 1) dy = 0 dA 1 = 0 and 8 − 6y − y 2 − y 4 dy = 2 4y − 1 A = 0 67 30 1 y (3 − y) − (y 2 + 1) dy ye dA = 1 3y 2 y3 y5 − − 2 6 10 0 (2y − y 2 − y 3 ) dy = y2 − 0 y3 y4 − 3 4 dA We can now ﬁnd the coordinates of the centroid. x= 1 A xe dA = A 67/30 67 = 7/6 35 and y= We conclude that the centroid is located at the following point. 67 5 , 35 14 (x, y) = 5 1 A ye dA = A 5/12 5 = 7/6 14 1 = 0 5 12 ...
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