Explanation Given information: The region is denoted as R . The region is bounded by a piecewise smooth simple closed curve C . The density is δ = 1 . The moments of inertia of R about the x and y axes are I x and I y respectively. Calculation: Write the equation. r = x 2 + y 2 (1) r 4 = ( x 2 + y 2 ) 4 = ( x 2 + y 2 ) 2 Take the vector field F = ∇ ( r 4 ) . F = ∇ ( r 4 ) = ∇ ( ( x 2 + y 2 ) 2 ) = ( 2 ( x 2 + y 2 ) ( 2 x ) ) i + ( 2 ( x 2 + y 2 ) ( 2 y ) ) j + ( 0 ) k = 4 x ( x 2 + y 2 ) i + 4 y ( x 2 + y 2 ) j ( F = M i + N j ) . M = 4 x ( x 2 + y 2 ) Differentiate M = 4 x ( x 2 + y 2 ) with respect to x . ∂ M ∂ x = 4 ( 1 ) ( x 2 + y 2 ) + ( 4 x ) ( 2 x + 0 ) = 4 ( x 2 + y 2 ) + 8 x 2 N = 4 y ( x 2 + y 2 ) Differentiate N = 4 y ( x 2 + y 2 ) with respect to y . ∂ N ∂ y = 4 ( 1 ) ( x 2 + y 2 ) + ( 4 y ) ( 0 + 2 y ) = 4 ( x 2 + y 2 ) + 8 y 2 The moments of inertia of R about the x -axis is I x = ∬ S ( y 2 + z 2 ) δ d σ = ∬ S y 2 d σ , since z = 0 and δ = 1

14th Edition

ISBN: 9780134438986

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 16.7, Problem 33E

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Evaluate the integral ∮C∇(r4)⋅nds with r=x2+y2 in terms of Ix and Iy.

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Chapter 16.7, Problem 32E

Chapter 16.7, Problem 34E

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Chapter 16 Solutions

Thomas' Calculus (14th Edition)

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Evaluate the integral ∮ C ∇ ( r 4 ) ⋅ n d s with r = x 2 + y 2 in terms of I x and I y .

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Evaluate the integral ∮C∇(r4)⋅nds with r=x2+y2 in terms of Ix and Iy.

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Chapter 16 Solutions