Explanation Given information: The curve function is y 2 = x 3 − x 4 (1) Calculation: Rearrange Equation (1) as shown below: y = ± x 3 − x 4 Therefore, the loop of the curve have the functions of y = x 3 − x 4 and y = − x 3 − x 4 . Draw the loop of the curve for the function y 2 = x 3 − x 4 using the procedure as shown below: Draw the graph for the function y = x 3 − x 4 by substituting different values for x . Similarly in the same graph plot for the function y = − x 3 − x 4 by substituting different values for x . The region enclosed by the curves y = x 3 − x 4 and y = − x 3 − x 4 is shown in Figure 1. Refer Figure 1. The loop of the curve is symmetric about x -axis. Therefore the y -coordinate of the centroid is y ¯ = 0 . The total area of the curve is twice the area of the top curve. The interval of the loop of the curve is x = 0 to x = 1 . The expression to find the area of the shaded region is shown below: A = ∫ a b [ f ( x ) ] d x (2) Here, the lower limit is a , the upper limit is b , and the top curve function is f ( x ) , and the bottom curve function is g ( x ) . Rearrange Equation (2) for the curve being symmetric. A = 2 ∫ a b [ f ( x ) ] d x (3) Substitute 0 for a , 1 for b , and x 3 − x 4 for f ( x ) in Equation (3). A = 2 ∫ 0 1 ( x 3 − x 4 ) d x = 2 ∫ 0 1 ( x 3 2 1 − x ) d x (4) Let sin θ = x (5) Find the limit of θ : Substitute 0 for x in Equation (5). sin θ = 0 θ = 0 Substitute 1 for x in Equation (5). sin θ = 1 sin θ = 1 θ = π 2 Substitute sin θ for x and change the limits as 0 to π 2 in Equation (4). A = 2 ∫ 0 π 2 ( 2 sin 4 θ cos θ 1 − sin 2 θ ) d θ = 2 ∫ 0 π 2 ( 2 sin 4 θ cos θ cos 2 θ ) d θ = 2 ∫ 0 π 2 ( 2 sin 4 θ cos 2 θ ) d θ = 4 ∫ 0 π 2 ( 1 4 ( 1 − cos 2 θ ) 2 1 2 ( 1 + cos 2 θ ) ) d θ = 1 2 ∫ 0 π 2 ( 1 − cos 2 θ − cos 2 2 θ + cos 3 2 θ ) d θ = 1 2 ∫ 0 π 2 ( 1 − cos 2 θ − 1 2 ( 1 + cos 4 θ ) + cos 2 θ ( 1 − sin 2 2 θ ) ) d θ = 1 2 ∫ 0 π 2 ( 1 − cos 2 θ − 1 2 − 1 2 cos 4 θ + cos 2 θ ( 1 − sin 2 2 θ ) ) d θ = 1 2 ∫ 0 π 2 ( 1 2 − 1 2 cos 4 θ − cos 2 θ sin 2 2 θ ) d θ (6) Integrate Equation (6) and apply the limits. A = 1 2 [ θ 2 − 1 8 sin 4 θ − 1 6 sin 3 2 θ ] 0 π 2 = 1 2 [ ( π 4 − 1 8 sin 4 ( π 2 ) − 1 6 sin 3 2 ( π 2 ) ) − ( 0 ) ] = 1 2 ( π 4 − 0 ) = π 8 Calculate the x -coordinate of the centroid ( x ¯ ) using the relation: x ¯ = 1 A ∫ a b x [ f ( x ) ] d x (7) Rearrange Equation (7) for the curve being symmetric

8th Edition

ISBN: 9781305266636

Author: James Stewart

Publisher: Cengage Learning

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Chapter 8, Problem 2P

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The centroid of the region enclosed by the curve y2=x3−x4 .

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The centroid of the region enclosed by the curve y 2 = x 3 − x 4 .

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