18.02SC Supplementary Problem Solutions: Triple Integrals

5. Triple Integrals 5A. Triple integrals in rectangular and cylindrical coordinates b) 2 1 0 � Middle: 1 2y � 5A-1 a) � 2 0 � √ 0 y � −1 � 1 (x + y + z)dx dy dz Inner: 0 + 21 y 2 + yz �1 y=−1 2xy 2 z dz dx dy Inner: xy 2 z 2 0 1 4 4 4x y + x(y + z) �√ y 0 = 41 y 6 �xy 0 �1 x=0 = 1 2 +y+z Outer: z + z 2 = 1 + z − (−z) = 1 + 2z xy Middle: 1 2 2x �2 0 =6 = x3 y 4 z 1 Outer: 1 7 28 y �2 0 = y+z=1 32 7 . 1 5A-2 a) (i) y � 0 1� 0 1� 1−y dz dy dx (ii) 0 � 0 1� 0 1−y � 1 dx dz dy (iii) 0 � 0 1� c) In cylindrical coordinates, with the polar coordinates r and θ in xz-plane, we get � π/2 � 1 � 2 ��� dy dr dθ = dy dr dθ R 0 0 0 1� 1−z 1 x dy dx dz 0 2 1 0 2 2 2 2 z 2 d) The sphere has equation x + y + z = 2, or r + z = 2 in cylindrical coordinates. The cone has equation z 2 = r2 , or z = r. The circle in which they intersect has a radius r found by solving the two equations z = r and z 2 +r2 = 2 simultaneously; eliminating z we get r2 = 1, so r = 1. Putting it all together, we get � 2π 0 � 1 0 � √ y 2−r 2 r dz dr dθ. cross-section view r 5A-3 By symmetry, x̄ = ȳ = z̄, so it suffices to calculate just one of these, say z̄. We have � 1 � 1−x � 1−x−y ��� 1 z dV = z dz dy dx z-moment = D 1 2 2z 0 0 �1−x−y �1−x 0 = 16 (1 − x)3 1 0 mass of D = volume of D = 13 (base)(height) = Therefore z̄ = 1 1 24 / 6 = 1 4; 1 3 z=1-x-y 0 = 12 (1 − x − y)2 Middle: − 61 (1 − x − y)3 0 �1 1 1 Outer: − 24 (1 − x)4 = 24 = z̄ moment. Inner: 2 1 · 1 2 · 1 = 16 . this is also x̄ and ȳ, by symmetry. 5A-4 Placing the cone as shown, its equation in cylindrical coordinates is z = r and the density is given by δ = r. By the geometry, its projection onto the xy-plane is the interior R of the origin-centered circle of radius h. y=1-x 1 z h h y vertical cross-section a) Mass of solid D = ��� Inner: (h − r)r2 ; D Middle: 2π h 0 � hr3 r4 − 3 4 �h δ dV = � 0 � h r · r dz dr dθ r = 0 h4 ; 12 Outer: 2πh4 12 b) By symmetry, the center of mass is on the z-axis, so we only have to compute its z-coordinate, z̄. � 2π � h � h ��� z δ dV = z-moment of D = zr · r dz dr dθ D 0 0 r � �h 1 2 r2 r5 1 2 = −r ) Middle: h − = h5 · Inner: 2 3 5 0 2 15 r 2 πh5 2πh5 4 . Therefore, z̄ = 15 Outer: = h. 2 4 15 5 πh 12 1 2 2 2z r �h 1 2 2 2 (h r 4 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.) The top of the tetrahedron is a plane intersecting the x- and y-axes at 1, and the z-axis at 2. Its equation is therefore x + y + 21 z = 1. The square of the distance of a point (x, y, z) to the axis of rotation (i.e., the x-axis) is given by y 2 + z 2 . We therefore get: � 1 � 1−x � 2(1−x−y) (y 2 + z 2 ) dz dy dx. moment of inertia = 4 0 0 0 5A-6 Placing D so its axis lies along the positive z-axis and its base is the � origin-centered disc √ of radius a in the xy-plane, the equation of the hemisphere is z = a2 − x2 − y 2 , or z = a2 − r2 in cylindrical coordinates. Doing the inner and outer integrals mentally: ��� � 2π � a � √a2 −r2 � a � 2 z-moment of inertia of D = r dV = r3 a2 − r2 dr. r2 dz r dr dθ = 2π D 0 0 0 0 √ The integral can be done using integration by parts (write the integrand r · r a2 − r2 ), or by substitution; following the latter course, we substitute r = a sin u, dr = a cos u du, and get (using the formulas at the beginning of exercises 3B) � a � � π/2 r3 a2 − r2 dr = a3 sin3 u · a2 cos2 u du 2 0 0 = a5 � 0 π/2 � � � � 2 2·4 2 5 sin3 u − sin5 u du = a5 = − a . 3 1·3·5 15 Ans: 4π 5 a . 15 5A-7 The solid D is bounded below by z = x2 + y 2 and above by z = 2x. The main problem is determining the projection R of D to the xy-plane, since we need to know this before we can put in the limits on the iterated integral. z The outline of R is the projection (i.e., vertical shadow) of the curve in which the paraboloid and plane intersect. This curve is made up of the points in which the graphs of z = 2x and z = x2 + y 2 intersect, i.e., the simultaneous solutions of the two equations. To project the curve, we omit the z-coordinates of the points on it. Algebraically, this amounts to solving the equations simultaneously by eliminating z from the two equations; doing this, we get as the outline of R the curve x2 + y 2 = 2x 4 D y x 2 cross-section of D view of D along x axis (x − 1)2 + y 2 = 1. or, completing the square, This is a circle of radius 1 and center at (1, 0), whose polar equation is therefore r = 2 cos θ. We use symmetry to calculate just the right half of D and double the answer: � π/2 � 2 cos θ � 2x r2 dz r dr dθ z-moment of inertia of D = 2 0 =2 � π/2 0 � 2 cos θ 0 � x2 +y 2 0 2r cos θ r3 dz dr dθ = 2 r2 � π/2 0 2 5 1 �2 cos θ 2 1 r cos θ − r6 = · 32 cos6 θ − · 32 cos6 θ 5 6 5 3 0 � π/2 32 1 · 3 · 5 π π 32 cos6 θ dθ = · · · = . Outer: · 15 0 15 2 · 4 · 6 2 3 � 2 cos θ r3 (2r cos θ−r2 ) dr dθ 0 Inner: 2π 3 Ans: 5B. Triple Integrals in spherical coordinates 5B-1 a) The angle between the central axis of the cone and any of the lines on√the cone is � 2π � π/4 � 2 √ π/4; the sphere is ρ = 2; so the limits are (no integrand given):: dρ dφ dθ. 0 b) The limits are (no integrand is given): � π/2 0 � π/2 0 � ∞ dρ dφ dθ 0 0 A 2 P 0 c) To get the equation of the sphere in spherical coordinates, we note that AOP is always a right triangle, for any position of P on the sphere. Since AO = 2 and OP = ρ, we get according to the definition of the cosine, cos φ = ρ/2, or ρ = 2 cos φ. (The picture shows the cross-section of the sphere by the plane containing P and the central axis AO.) 1 φ O cross-section The plane z = 1 has in spherical coordinates the equation ρ cos φ = 1, or ρ = sec φ. It intersects the sphere in a circle of radius 1; this shows that π/4 is the maximum value of φ for which the ray having angle φ intersects the region.. Therefore the limits are (no integrand is given): � 2π � π/4 � 2 cos φ dρ dφ dθ. 0 0 sec φ z 5B-2 Place the solid hemisphere D so that its central axis lies along the positive z-axis and its base is in the xy-plane. By symmetry, x̄ = 0 and ȳ = 0, so we only need z̄. The integral for it is the product of three separate one-variable integrals, since the integrand is the product of three one-variable functions and the limits of integration are all constants. ��� � 2π � π/2 � a (ρ cos φ)ρ2 sin φ dρ dφ dθ z̄-moment = z dV = D 0 0 = 2π · Since the mass is � 0 �a � �π/2 1 4 1 πa4 1 1 ρ sin2 φ = . · = 2π · a4 · 4 2 4 2 4 0 0 2 3 πa4 /4 3 πa , we have finally z̄ = = a. 3 2πa3 /3 8 5B-3 Place the solid so the vertex is at the origin, and the central axis lies along the positive z-axis. In spherical coordinates, the density is given by δ = z = ρ cos φ, and referring to the picture, we have ��� ��� M. of I. = r2 · z dV = (ρ sin φ)2 (ρ cos φ) ρ2 sin φ dρ dφ dθ D a D = � 2π 0 � π/6 0 � a z a r ρ π φ 6 y ρ5 sin3 φ cos φ dρ dφ dθ cross-section 0 � �4 �π/6 πa6 a6 1 1 a 1 · sin4 φ · = 6 . = 2π · 6 4 6 4 2 2 ·3 0 6 = 2π · 5B-4 Place the sphere so its center is at the origin. In each case the iterated integral can be expressed as the product of three one-variable integrals (which are easily calculated), since the integrand is the product of one-variable functions and the limits are constants. � 2π � π � a 1 πa4 3a ρ · ρ2 sin φ dρ dφ dθ = 2π · 2 · a4 = πa4 ; a) average = = . 3 4 4πa /3 4 0 0 0 b) Use the z-axis as diameter. The distance of a point from the z-axis is r = ρ sin φ. � 2π � π � a π 1 π 2 a4 π 2 a4 /4 3πa ρ sin φ · ρ2 sin φ dρ dφ dθ = 2π · · a4 = ; average = = . 3 2 4 4 4πa /3 16 0 0 0 c) Use the xy-plane and the upper solid hemisphere. The distance is z = ρ cos φ. � 2π � π/2 � a πa4 πa4 /4 3a 1 1 ; average = = . ρ cos φ · ρ2 sin φ dρ dφ dθ = 2π · · a4 = 3 /3 2 4 4 2πa 8 0 0 0 5C. Gravitational Attraction 1 5C-2 The top of the cone is given by z = 2; in spherical coordinates: ρ cos φ = 2. Let α be the angle between the axis of the cone and any of its generators. The density δ = 1. Since the cone is symmetric about its axis, the gravitational attraction has only a k-component, and is � 2π � α � 2/ cos φ sin φ cos φ dρ dφ dθ. G 0 Inner: 2 sin φ cos φ cos φ 0 2 5 α 0 Middle: −2 cos φ �α 0 = −2 cos α + 2 Outer: 2π · 2(1 − cos α) � � 2 Ans: 4πG 1 − √ . 5 5C-3 Place the sphere as shown so that Q is at the origin. Since it is rotationally symmetric about the z-axis, the force will be in the k -direction. Equation of sphere: ρ = 2 cos φ Density: δ = ρ−1/2 � 2π � π/2 � 2 cos φ z ρ−1/2 cos φ sin φ dρ dφ dθ Fz = G 0 0 0 �2 cos φ √ Inner: cos φ sin φ 2ρ1/2 = 2 2 cos3/2 φ sin φ 0 √ √ √ �π/2 � √ 4 2 8 2 4 2 2 5/2 = Outer: 2πG = π G. Middle: 2 2 − cos φ 5 5 5 5 0 Q 5C-4 Referring to the figure, the total gravitational attraction (which is in the k direction, by rotational symmetry) is the sum of the two integrals G � 2π 0 � π/3 0 � 1 cos φ sin φ dρ dφ dθ 0 1 = 2π G · 2 + G � 2π 0 � π/2 π/3 � 2 cos φ cos φ sin φ dρ dφ dθ 0 � √ �2 � �3 3 2 1 3 1 11 + 2π G · = πG+ πG = π G. 2 3 2 4 6 12 z The two spheres are shown in cross-section. The spheres inter­ sect at the points where φ = π/3. The first integral respresents the gravitational attraction of the top part of the solid, bounded below by the cone φ = π/3 and above by the sphere ρ = 1. The second integral represents the bottom part of the solid, bounded below by the sphere ρ = 2 cos φ and above by the cone. π 3 y y