Explanation Given information: r = cos 2 θ , ρ ( x , y ) = x 2 + y 2 . Formula used: Use maple x ¯ = 1 m ∫ D x x 2 + y 2 d A . Calculation: Consider r = cos 2 θ , ρ ( x , y ) = x 2 + y 2 . The integral for the mass is, ∬ D x 2 + y 2 d A . Convert to polar by substituting r cos θ for x and r sin θ for y , integrating across 0 ≤ r ≤ cos 2 θ and − π 4 ≤ θ ≤ π 4 . Then ∫ − π 4 π 4 ∫ 0 cos 2 θ ( r sin θ ) 2 + ( r cos θ ) 2 r d r d θ = ∫ − π 4 π 4 ∫ 0 cos 2 θ r 2 sin 2 θ + cos 2 θ r d r d θ = ∫ − π 4 π 4 ∫ 0 cos 2 θ r 2 r d r d θ Use maple to evaluate ∫ − π 4 π 4 ∫ 0 cos 2 θ r 2 r d r d θ . Then > ∫ − π 4 π 4 ∫ 0 2 ( cos ( θ ) ) 2 − 1 r 3 d r d θ 3 π 64 Therefore, the mass is 3 π 64 . we know that, x ¯ = 1 m ∫ D x x 2 + y 2 d A Convert to polar by substituting r cos θ for x and r sin θ for y , integrating across 0 ≤ r ≤ cos 2 θ and − π 4 ≤ θ ≤ π 4 . Also substitute 3 π 64 for m . 1 3 π 64 ∫ − π 4 π 4 ∫ 0 cos 2 θ r cos θ ( r cos θ ) 2 + ( r sin θ ) 2 r d r d θ = 64 3 π ∫ − π 4 π 4 ∫ 0 cos 2 θ r 4 cos θ d r d θ Enter 64 3 π ∫ − π 4 π 4 ∫ 0 cos 2 θ r 4 cos θ d r d θ in maple. > 64 3 π ∫ − π 4 π 4 ∫ 0 2 ( cos ( θ ) ) 2 − 1 r 4 cos ( θ ) d r d θ 16384 10395 2 π Thus, 64 3 π ∫ − π 4 π 4 ∫ 0 cos 2 θ r 4 cos θ d r d θ = 16384 2 10395 π We know that, y ¯ = 1 m ∫ D y x 2 + y 2 d A . Convert to polar by substituting r cos θ for x and r sin θ for y , integrating across 0 ≤ r ≤ cos 2 θ and − π 4 ≤ θ ≤ π 4 . Also substitute 3 π 64 for m . 1 3 π 64 ∫ − π 4 π 4 ∫ 0 cos 2 θ r sin θ ( r cos θ ) 2 + ( r sin θ ) 2 r d r d θ = 64 3 π ∫ − π 4 π 4 ∫ 0 cos 2 θ r 4 sin θ d r d θ Enter 64 3 π ∫ − π 4 π 4 ∫ 0 cos 2 θ r 4 sin θ d r d θ in maple. > 64 3 π ∫ − π 4 π 4 ∫ 0 2 ( cos ( θ ) ) 2 − 1 r 4 sin ( θ ) d r d θ Thus, 64 3 π ∫ − π 4 π 4 ∫ 0 cos 2 θ r 4 sin θ d r d θ = 0 Therefore, the center of mass is ( x ¯ , y ¯ ) = 16384 2 10395 π , 0

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ISBN: 9780357687901

Author: Stewart

Publisher: CENGAGE L

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Chapter 15.4, Problem 27E

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To find: the mass, center of mass, and moments of inertia of the lamina.

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Chapter 15.4, Problem 26E

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Chapter 15 Solutions

EBK CALCULUS EARLY TRANSCENDENTALS

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the mass, center of mass, and moments of inertia of the lamina.

Solution Summary: The author explains how to find the mass, center of mass and moments of inertia of the lamina.

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To find: the mass, center of mass, and moments of inertia of the lamina.

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Chapter 15 Solutions