5. Triple Integrals Solutions

5. Triple Integrals Solutions - 5 Triple Integrals 5A...

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5. Triple Integrals 5A. Triple integrals in rectangular and cylindrical coordinates 1 Middle: iy + iy2 + yz ] = 1 + z - (-1) = 1 + 2z Outer: z + z2] 2 = 6 y=-1 0 X Y dz dx dy Inner: xy2z2] o = x3y4 Jd2 Jdfi JdxY b) ~ X Y ~ Z Middle: ,x 2 1 4 y 4 ] f i = iy6 Outer: kY7] = y. 0 5A-2 dz dy dx (ii) Jdl Jdl-' Jdl dxdzdy (iii)JdlJdlJdl-'dydXdz c) In cylindrical coordinates, with the polar coordinates r and 8 in xz-plane, we get 11, dy dr de = Jdr" Jdl Jd2 dy dr do d) The sphere has equation x2 + y2 + z2 = 2, or r2 + z2 = 2 in cylindrical coordinates. The cone has equation z2 = r2, or z = r. The circle in which they intersect has a radius r found by solving the two equations z = r and z2+r2 = 2 simultaneously; .Y eliminating z we get r2 = 1, so r = 1. Putting it all together, we get \ : I cross-sectionview 5A-3 By symmetry, i = y = 2, so it suffices to calculate just one of these, say 2. We have 1-x 1-z-y z-moment = z d v = z dz dy dx z= 1-X-y Jdl Jd Jd 1-x-y 1-2 1nner: iz2] = ; ( l - ~ - y ) ~ M i d d l e : - i ( l - ~ - y ) ~ ] ~ = i ( l - ~ ) ~ 0 1 Outer: - 1 - x 4 ] = I - z moment. - ,, 0 v= 1-x mass of D = volume of D = i(base)(height) = i - ; . 1 = i . Therefore 2 = &/+ = i; this is also Z and g, by symmetry. 5A-4 Placing the cone as shown, its equation in cylindrical coordinates is z = r and the density is given by 6 = r. By the geometry, its projection onto the xy-plane is the interior R of the origin-centered circle of radius h. vertical cross-section

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TRIPLE INTEGRALS a) Mass of solid D = / ~ ~ ~ 6 d V = l ~ ~ l ~ ~ l ~ ~ . r d z d r d 6 2 ~ h 4 Inner: (h - r)r2; Middle: - - - = . Outer: - 3 hr3 4 , 12' 12 r41h b) By symmetry, the center of mass is on the z-axis, so we only have to compute its z-coordinate, 5. z-moment of D = / / L z 6 d v = ~ 2 r ~ h J l h z T ~ T d z d T d ~ h 1 Inner: iz2r2] = i(h2r2 - r4) Middle: g) = -h5 . - 2 2 A (h2: 2 15 - T , 5A-5 Position S so that its base is in the xy-plane and its diagonal D lies along the x-axis (the y-axis would do equally well). The octants divide S into four tetrahedra, which by symmetry have the same moment of inertia about the x-axis; we calculate the one in the first octant. (The picture looks like that for 5A-3, except the height is 2.)
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