Explanation Given: The given inequalities y ≤ 1 4 x 2 , ( x − 4 ) 2 + y 2 ≤ 16 , y ≥ 0. Formula used: Centroid of a region ( x ¯ , y ¯ ) bounded by two curves f ( x ) and g ( x ) in the interval [ a , b ] is, x ¯ = 1 A ( ∫ a b x ( f ( x ) − g ( x ) ) d x ) , and, y ¯ = 1 A ∫ a b 1 2 ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x A is the area of the bounded region. Calculation: Consider the following inequalities, y ≤ 1 4 x 2 , ( x − 4 ) 2 + y 2 ≤ 16 , y ≥ 0. y ≤ 1 4 x 2 is the region in anupward parabola with vertex at ( 0 , 0 ) . Whereas, ( x − 4 ) 2 + y 2 ≤ 16 is the region within the circle, with center at ( 4 , 0 ) and radius 4. Also, y ≥ 0 is the region above x -axis. Now, the point of intersection of circle and the parabola is to calculate. We consider the following equations, y = 1 4 x 2 and ( x − 4 ) 2 + y 2 = 16 Tosolve the equation, substitute y = 1 4 x 2 in the equation ( x − 4 ) 2 + y 2 = 16 . Therefore, ( x − 4 ) 2 + 1 16 x 4 = 16 x 2 + 16 − 8 x + 1 16 x 4 = 16 x 4 + 16 x 2 − 128 x = 0 x ( x − 4 ) ( x 2 + 4 x + 32 ) = 0 Because, ( x 2 + 4 x + 32 ) do not have any real solution either x = 0 or x = 4 . So, substitute x = 0 then, y = 0 . when x = 4 , then y = 4 . Thus, the graph of the inequalities with the common region is as following, It is clear from the graph thatarea of the shaded region, which is equal to the area under the parabolic curve y = 1 4 x 2 in the interval [ 0 , 4 ] and area of the quarter of the circle with radius 4. Thus, Area of the shaded region = ∫ 0 4 1 4 x 2 d x + π ( 4 ) 2 4 = [ 1 12 x 3 ] 0 4 + 4 π = [ 1 12 ( 4 ) 3 − 1 12 ( 0 ) 3 ] + 4 π = 16 3 + 4 π …… (1) Consider the following equation, ( x − 4 ) 2 + y 2 = 16 . ( x − 4 ) 2 + y 2 = 16 is equivalent to y = ± 16 − ( x − 4 ) 2 . It is clear from the parabola that, the common region is only above the x- axis, so consider only the equation, y = 16 − ( x − 4 ) 2 . Let assume f ( x ) = 1 4 x 2 and g ( x ) = 16 − ( x − 4 ) 2 Calculate ∫ a b x ( f ( x ) − g ( x ) ) d x for the x ¯ of the centroid,. By taking the upper limit as 8 and the lower limit as 4 of the function g ( x ) and change the sign of g ( x ) different limits are taken. x ¯ = 1 A ( ∫ a b x ( f ( x ) − g ( x ) ) d x ) is equivalent to x ¯ = 1 A ( ∫ a b x ⋅ f ( x ) d x + ∫ c d x ⋅ g ( x ) d x ) Thus, ∫ a b x ( f ( x ) − g ( x ) ) d x = ∫ a b x ⋅ f ( x ) d x + ∫ c d x ⋅ g ( x ) d x …… (2) Hence, ∫ a b x ⋅ f ( x ) d x = ∫ 0 4 x ⋅ 1 4 x 2 d x = [ 1 16 x 4 ] 0 4 = [ 1 16 ( 4 ) 4 − 1 16 ( 4 ) 0 ] = 16 …… (3) So, ∫ 4 8 x ⋅ g ( x ) d x = ∫ 4 8 x ⋅ 16 − ( x − 4 ) 2 d x = ∫ 4 8 ( x − 4 ) ⋅ 16 − ( x − 4 ) 2 d x + ∫ 4 8 4 ⋅ 16 − ( x − 4 ) 2 d x = ∫ 4 8 ( x − 4 ) ⋅ 16 − ( x − 4 ) 2 d x + ∫ 4 8 4 ⋅ 4 2 − ( x − 4 ) 2 d x …… (4) Now, substitute 16 − ( x − 4 ) 2 = t in ∫ 4 8 ( x − 4 ) ⋅ 16 − ( x − 4 ) 2 d x . Then, ∫ 4 8 ( x − 4 ) ⋅ 16 − ( x − 4 ) 2 d x = ∫ 4 8 ( x − 4 ) t d t − 2 ( x − 4 ) = − 1 2 [ t 3 2 ] 4 8 Substitute t = 16 − ( x − 4 ) 2 to get, − 1 2 [ t 3 2 ] 4 8 = − 1 2 [ ( 16 − ( x − 4 ) 2 ) 3 2 ] 4 8 = − 1 2 [ ( 16 − ( 8 − 4 ) 2 ) 3 2 − ( 16 − ( 4 − 4 ) 2 ) 3 2 ] = 1 2 ( 16 3 2 ) Now, ∫ 4 8 ( x − 4 ) ⋅ 16 − ( x − 4 ) 2 d x = 1 2 ( 16 3 2 ) …… (5) By using ∫ ( a 2 − x 2 ) d x = x a 2 − x 2 2 + a 2 2 arcsin ( x a ) in ∫ 4 8 4 ⋅ 4 2 − ( x − 4 ) 2 d x

Student Solutions Manual For Larson/edwards' Calculus Of A Single Variable: Early Transcendental Functions, 2nd

7th Edition

ISBN: 9781337552561

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 8.4, Problem 58E

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To calculate: The centroid of the region bounded by the graphs of the following inequalities,

y≤14x2,(x−4)2+y2≤16,y≥0.

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Chapter 8.4, Problem 57E

Chapter 8.4, Problem 59E

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Chapter 8 Solutions

Student Solutions Manual For Larson/edwards' Calculus Of A Single Variable: Early Transcendental Functions, 2nd

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The centroid of the region bounded by the graphs of the following inequalities, y ≤ 1 4 x 2 , ( x − 4 ) 2 + y 2 ≤ 16 , y ≥ 0.

Solution Summary: The author calculates the centroid of a region bounded by the graphs of the following inequalities.

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To calculate: The centroid of the region bounded by the graphs of the following inequalities,

y≤14x2,(x−4)2+y2≤16,y≥0.

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Chapter 8 Solutions

The centroid of the region bounded by the graphs of the following inequalities, y ≤ 1 4 x 2 , ( x − 4 ) 2 + y 2 ≤ 16 , y ≥ 0.

Solution Summary: The author calculates the centroid of a region bounded by the graphs of the following inequalities.

Concept explainers

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To calculate: The centroid of the region bounded by the graphs of the following inequalities, y≤14x2,(x−4)2+y2≤16,y≥0.

Want to see the full answer?

Chapter 8 Solutions

To calculate: The centroid of the region bounded by the graphs of the following inequalities,

y≤14x2,(x−4)2+y2≤16,y≥0.