5. Triple Integrals. 5A. Triple integrals in rectangular and cylindrical coordinates. 2 + y + z x=0. y Outer: 1

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1 5. Triple Integrals 5A. Triple integrals in rectangular and clindrical coordinates ] 5A- a) (x + + )dxdd Inner: x + x( + ) + + x ] ] Middle: ( ) + Outer: + 6 x ] x b) x ddxd Inner: x x 3 4 ] ] + Middle: 4 x Outer: A- a) (i) dddx (ii) dxdd (iii) ddxd x c) In clindrical coordinates, with the polar coordinates r and θ in x-plane, we get π/ ddrdθ ddrdθ R d) The sphere has equation x + +, or r + in clindrical coordinates. The cone has equation r, or r. The circle in which the intersect has aradius r foundbsolvingthetwoequations r and +r simultaneousl; eliminating we get r, so r. Putting it all together, we get π r rddrdθ. r cross-section view 5A-3 B smmetr, x ȳ, so it suffices to calculate just one of these, sa. We have x x -moment dv dddx -x- D ] x ] x Inner: ( x ) Middle: 6 ( x )3 6 ( x)3 Outer: ( x) 4 ] moment. 4 4 mass of D volume of D 3(base)(height) 3 Therefore / ; this is also x and ȳ, b smmetr A-4 Placing the cone as shown, its equation in clindrical coordinates is r and the densit is given b δ r. B the geometr, its projection onto the x-plane is the interior R of the origin-centered circle of radius h. 6. -x h h vertical cross-section

2 π h h a) Mass of solid D δdv r rddrdθ TRIPLE INTEGRALS D r hr 3 r 4 ] h h 4 πh 4 Inner: (h r)r ; Middle: 3 4 ; Outer: b) B smmetr, the center of mass is on the -axis, so we onl have to compute its -coordinate,. π h h -moment of D δdv r rddrdθ D r ] ( h h r r 5 ) h Inner: h 5 r (h r r 4 ) Middle: r πh 5 πh Outer:. Therefore, h. 5 πh 4 5 5A-5 Position S so that its base is in the x-plane and its diagonal D lies along the x-axis (the -axis would do equall well). The octants divide S into four tetrahedra, which b smmetr 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.) The top of the tetrahedron is a plane intersecting the x- and -axes at, and the -axis at. Its equation is therefore x + +. The square of the distance of a point (x,,) to the axis of rotation (i.e., the x-axis) is given b +. We therefore get: x ( x ) moment of inertia 4 ( + )dddx. 5A-6 Placing D so its axis lies along the positive -axis and its base is the origin-centered disc of radius a in the x-plane, the equation of the hemisphere is a x, or a r in clindrical coordinates. Doing the inner and outer integrals mentall: π a a r a 3 -moment ofinertiaof D r dv r drdrdθ π r a r dr. D The integral can be done using integration b parts (write the integrand r r a r ), or b substitution; following the latter course, we substitute r asinu, dr acosudu, and get (using the formulas at the beginning of exercises 3B) a π/ 3 r a r 3 dr a sin 3 u a cos udu π/ ( ) ( 5 a sin 3 u sin 5 ) π 5 u du a a. Ans: a

3 E. 8. EXERCISES 5A-7 The solid D is bounded below b x + and above b x. The main problem is determining the projection R of D to the x-plane, since we need to know this before we can put in the limits on the iterated integral. The outline of R is the projection (i.e., vertical shadow) of the curve 4 in which the paraboloid and plane intersect. This curve is made up of the points in which the graphs of x and x + intersect, D i.e., the simultaneous solutions of the two equations. To project the curve, we omit the -coordinates of the points on it. Algebraicall, this x amounts to solving the equations simultaneousl b eliminating from the two equations; doing this, we get as the outline of R the curve cross-section of D view of D along x axis x + x or, completing the square, (x ) +. This is a circle of radius and center at (,), whose polar equation is therefore r cosθ. We use smmetr to calculate just the right half of D and double the answer: π/ cos θ x -moment of inertia of D r drdrdθ x + π/ cos θ r cos θ π/ cos θ 3 r ddrdθ r 3 (r cosθ r )drdθ r 5 ] cos θ 6 Inner: r cosθ r 3cos 6 θ 3cos 6 θ π/ π π π Outer: cos θdθ. Ans: B. Triple Integrals in spherical coordinates 5B- a) The angle between the central axis of the cone and an of the lines on the cone is π π/4 π/4; the sphere is ρ ; so the limits are (no integrand given):: dρdφdθ. π/ π/ A b) The limits are (no integrand is given): dρdφdθ P c) To get the equation of the sphere in spherical coordinates, we note that AOP is alwas a right triangle, for an position of P on the sphere. Since AO and OP ρ, we get according to the definition of the cosine, cosφ ρ/, or ρ cosφ. (The picture shows the cross-section of the sphere b the plane containing P and the central axis AO.) φ O cross-section The plane has in spherical coordinates the equation ρcosφ, or ρ secφ. It intersects the sphere in a circle of radius ; this shows that π/4 is the maximum value of φ for which the ra having angle φ intersects the region.. Therefore the limits are (no integrand is given): π π/4 cos φ dρdφdθ. sec φ

4 TRIPLE INTEGRALS 3 5B- Place the solid hemisphere D so that its central axis lies along the positive -axis and its base is in the x-plane. B smmetr, x and ȳ, so we onl need. 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. π π/ a -moment dv (ρcosφ)ρ sinφdρdφdθ D ( ) a ( ) π/ ρ 4 4 πa 4 π sin φ π a Since the mass is πa 3 πa 4 /4, we have finall 3 a. 3 πa 3 /3 8 5B-3 Place the solid so the vertex is at the origin, and the central axis lies along the positive -axis. In spherical coordinates, the densit is given b δ ρcosφ, and referring to the picture, we have M. of I. r dv (ρsinφ) (ρcosφ)ρ sinφdρdφdθ D D a r π π/6 a ρ 5 sin 3 φcosφdρdφdθ cross-section ( a 6 ] ) π/6 a 6 4 πa 6 π sin 4 φ π a π 6 ρ φ 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 easil calculated), since the integrand is the product of one-variable functions and the limits are constants. π π a πa 4 3a a) ρ ρ 4 sinφdρdφdθ π a πa 4 ; average. 4 4πa 3 /3 4 b) Use the -axis as diameter. The distance of a point from the -axis is r ρsinφ. π π a π 4 π a 4 π a 4 /4 3πa ρsinφ ρ sinφdρdφdθ π a ; average πa 3 /3 6 c) Use the x-plane and the upper solid hemisphere. The distance is ρcosφ. π π/ a 4 πa 4 πa 4 /4 3a ρcosφ ρ sinφdρdφdθ π a ; average. 4 4 πa 3 /3 8

5 4 E. 8. EXERCISES 5C. Gravitational Attraction 5C- The top of the cone is given b ; in spherical coordinates: ρcosφ. Let α be the angle between the axis of the cone and an of its generators. The densit δ. Since the cone is smmetric about its axis, the gravitational attraction has onl a k-component, and is α 5 π α /cos φ G sinφcosφdρdφdθ. ] α Inner: sinφcosφ Middle: cosφ cosα+ Outer: π ( cosα) cosφ ( ) Ans: 4πG. 5 5C-3 Place the sphere as shown so that Q is at the origin. Since it is rotationall smmetric about the -axis, the force will be in the k -direction. Equation of sphere: ρ cosφ Densit: δ ρ / π π/ cos φ F G ρ / cosφsinφdρdφdθ ] cos φ Inner: cosφsinφ ρ / cos 3/ φ sinφ [ ] π/ / Middle: cos φ Outer: πg πg Q 5C-4 Referring to the figure, the total gravitational attraction (which is in the k direction, b rotational smmetr) is the sum of the two integrals π π/3 π π/ cos φ G cosφ sinφdρdφdθ + G cosφ sinφdρdφdθ π/3 ( ) ( ) πg + πg πg+ πg πg The two spheres are shown in cross-section. The spheres intersect at the points where φ π/3. The first integral respresents the gravitational attraction of the top part of the solid, bounded below b the cone φ π/3 and above b the sphere ρ. The second integral represents the bottom part of the solid, bounded below b the sphere ρ cosφ and above b the cone. π 3

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