Explanation Formula used: The moment of inertia is expressed mathematically as I y = ρ ∫ 0 a x 2 y d x . The mass is given by the equation m = ρ ∫ 0 a y 2 d x Calculation: The given equation is x 2 + y 2 = a 2 . Rewrite the given equation as shown below: x 2 + y 2 = a 2 y 2 = a 2 − x 2 y = a 2 − x 2 Substitute y = a 2 − x 2 , in I y = ρ ∫ 0 a x 2 y d x and obtain the value of the area as follows: I y = ρ ∫ 0 a x 2 y d x = ρ ∫ 0 a x 2 a 2 − x 2 d x Let x = a sin θ , then d x = a cos θ d θ and a 2 − x 2 = a cos θ ⋯ ⋯ ( 1 ) . Also, note that θ = sin − 1 x a The above equation can be rewritten as shown below: I y = ρ ∫ 0 a x 2 a 2 − x 2 d x = ρ ∫ 0 a ( a sin θ ) 2 a 2 − ( a sin θ ) 2 a cos θ d θ = ρ ∫ 0 a a 2 sin 2 θ a 1 − sin 2 θ a cos θ d θ = ρ ∫ 0 a a 4 sin 2 θ cos 2 θ cos θ d θ = ρ ∫ 0 a a 4 sin 2 θ cos 2 θ d θ Simplify further as follows: I y = ρ ∫ 0 a a 4 sin 2 θ cos 2 θ d θ = ρ ∫ 0 a a 4 sin 2 θ ( 1 − sin 2 θ ) d θ ( ∵ 1 − sin 2 θ = cos 2 θ ) = a 4 ρ ∫ 0 a ( sin 2 θ − sin 4 θ ) d θ = ρ a 4 [ ∫ 0 a ( sin 2 θ ) d θ − ∫ 0 a ( sin 4 θ ) d θ ] = ρ a 4 [ ∫ 0 a ( sin 2 θ ) d θ − ∫ 0 a ( sin 4 θ ) d θ ] Simplify further the above equation as follows: I y = ρ a 4 [ ∫ 0 a ( sin 2 θ ) d θ − ∫ 0 a ( sin 4 θ ) d θ ] = ρ a 4 [ ∫ 0 a 1 2 ( 1 − cos 2 θ ) d θ − 1 4 ∫ 0 a ( 1 − cos θ ) 2 d θ ] ( ∵ sin 2 θ = 1 2 ( 1 − cos 2 θ ) sin 4 θ = 1 4 ( 1 − cos θ ) 2 ) = ρ a 4 [ 1 2 ∫ 0 a d θ − 1 2 ∫ 0 a cos 2 θ d θ − 1 4 ∫ 0 a ( 1 − 2 cos θ + cos 2 θ ) d θ ] = ρ a 4 [ 1 2 ∫ 0 a d θ − 1 2 ∫ 0 a cos 2 θ d θ − 1 4 ∫ 0 a d θ + 1 2 ∫ 0 a ( cos θ ) d θ − 1 4 ∫ 0 a ( cos 2 2 θ ) d θ ] = ρ a 4 [ 1 2 ∫ 0 a d θ − 1 4 ∫ 0 a d θ − 1 4 ∫ 0 a ( 1 + cos 4 θ 2 ) d θ ] ( cos 2 2 θ = 1 + cos 2 θ 2 ) The above equation can be further simplified as shown below: I y = ρ a 4 [ 1 2 ∫ 0 a d θ − 1 4 ∫ 0 a d θ − 1 4 ∫ 0 a ( 1 + cos 4 θ 2 ) d θ ] = ρ a 4 [ 1 4 ∫ 0 a d θ − 1 8 ∫ 0 a d θ − 1 8 ∫ 0 a cos 4 θ d θ ] = ρ a 4 [ 1 4 θ − 1 8 θ − 1 32 sin 4 θ ] = ρ a 4 [ 1 8 θ − 1 32 sin 4 θ ] 0 a It is known that sin 4 θ can be written as shown below: sin 4 θ = 2 sin θ cos 2 θ = 2 [ 2 sin θ cos θ ( 1 − 2 sin 2 θ ) ] ( ∵ cos 2 θ = 1 − 2 sin 2 θ ) = 4 sin θ cos θ ( 1 − 2 sin 2 θ ) Substitute sin 4 θ = 2 [ 2 sin θ cos θ ( 1 − 2 sin 2 θ ) ] in the above equation and simplify

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Chapter 28.8, Problem 31E

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The moment of inertia of the flat plate with respect to y-axis in terms of its mass.

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Chapter 28.8, Problem 30E

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Chapter 28 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

Ch. 28.1 - Integrate: . Ch. 28.1 - Prob. 2PE Ch. 28.1 - Prob. 1E Ch. 28.1 - In Exercises 1 and 2, make the given changes in...Ch. 28.1 - Prob. 3E Ch. 28.1 - Prob. 4E Ch. 28.1 - Prob. 5E Ch. 28.1 - Prob. 6E Ch. 28.1 - Prob. 7E Ch. 28.1 - Prob. 8E

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The moment of inertia of the flat plate with respect to y -axis in terms of its mass.

The moment of inertia of the flat plate with respect to y-axis in terms of its mass.

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Chapter 28 Solutions