# 2 cos 2 θ dθ 1 6 δ 10 θ 3 sin 2 θ 2 π 0 10 3

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+ 2 cos 2 θ = 1 6 δ 10 θ + 3 sin 2 θ 2 π 0 = 10 3 πδ. Because the mass m of S is the product of its surface area and its density, we have m = 4 πδ , and hence we may also express I z in the form I z = 5 6 m = m · (0 . 9128709291752769) 2 . 14.5.10: The surface S has equation z = h ( x, y ) = x 2 + y 2 and lies over the annular region R in the xy -plane described in polar coordinates by 2 r 5, 0 θ 2 π . We have surface area element dS = 1 + x 2 x 2 + y 2 + y 2 x 2 + y 2 1 / 2 dA = 2 dA, and therefore the moment of inertia of the constant-density surface S with respect to the z -axis is I z = S ( x 2 + y 2 ) δ dS = 2 π 0 5 2 δr 3 2 dr dθ = δ 2 2 π 0 1 4 r 4 5 2 = 609 2 πδ 2 . The mass of S is m = S δ dS = 2 π 0 5 2 δr 2 dr dθ = 21 δπ 2 , and therefore its moment of inertia with respect to the z -axis may also be expressed in the form I z = 29 2 m m · (3 . 8078865529319541) 2 . 14.5.11: The surface S has the spherical-coordinates parametrization r ( φ, θ ) = 5 sin φ cos θ, 5 sin φ sin θ, 5 cos φ , 0 φ arccos 3 5 , 0 θ 2 π. 1872
Therefore r φ × r θ = i j k 5 cos φ cos θ 5 cos φ sin θ 5 sin φ 5 sin φ sin θ 5 sin φ cos θ 0 = 25 sin 2 φ cos θ, 25 sin 2 φ sin θ, 25 sin φ cos φ , and thus | r φ × r θ | = 625 sin 2 φ cos 2 φ + 625 sin 4 φ cos 2 θ + 625 sin 4 φ sin 2 θ = 25 sin φ (because 0 φ π/ 2). Hence the mass of S is m = S δ dS = 2 π θ =0 arccos(3 / 5) φ =0 25 δ sin φ dφ dθ = 2 πδ 25 cos φ arccos(3 / 5) 0 = 20 πδ (62 . 8318530717958648) δ. Next, x 2 + y 2 = (5 sin φ cos θ ) 2 + (5 sin φ sin θ ) 2 = 25 sin 2 φ, and hence the moment of inertia of S with respect to the z -axis is I z = S ( x 2 + y 2 ) δ dS = 2 π θ =0 arccos(3 / 5) φ =0 625 δ sin 3 φ dφ dθ = 2 πδ 625 3 cos 3 φ 625 cos φ arccos(3 / 5) 0 = 520 3 πδ (544 . 5427266222308280) δ. The moment of inertia may also be expressed in the form I z = 26 3 m m · (2 . 9439202887759490) 2 . 14.5.12: The upper half of the surface S has the spherical-coordinates parametrization r ( φ, θ ) = 5 sin φ cos θ, 5 sin φ sin θ, 5 cos φ , arccos 4 5 φ 1 2 π, 0 θ 2 π. To find the mass and moment of inertia with respect to the z -axis, we will integrate over the top half of S and then double the result. But first, r φ × r θ = i j k 5 cos φ cos θ 5 cos φ sin θ 5 sin φ 5 sin φ sin θ 5 sin φ cos θ 0 = 25 sin 2 φ cos θ, 25 sin 2 φ sin θ, 25 sin φ cos φ , and thus | r φ × r θ | = 625 sin 2 φ cos 2 φ + 625 sin 4 φ cos 2 θ + 625 sin 4 φ sin 2 θ = 25 sin φ 1873
(because 0 φ π/ 2). Hence the mass of S is m = S δ dS = 2 2 π θ =0 π/ 2 φ =arccos(4 / 5) 25 δ sin φ dφ dθ = 4 πδ 25 cos φ π/ 2 arccos(4 / 5) = 80 πδ (251 . 3274122871834591) δ. Next, x 2 + y 2 = (5 sin φ cos θ ) 2 + (5 sin φ sin θ ) 2 = 25 sin 2 φ, and hence the moment of inertia of S with respect to the z -axis is I z = S ( x 2 + y 2 ) δ dS = 2 2 π θ =0 π/ 2 φ =arccos(4 / 5) 625 δ sin 3 φ dφ dθ = 4 πδ 625 3 cos 3 φ 625 cos φ π/ 2 arccos(4 / 5) = 4720 3 πδ (4942 . 7724416479413618) δ. The moment of inertia may also be expressed in the form I z = 59 3 m m · (4 . 4347115652166902) 2 .