) in the k-th subbox. The mass of the k-th subbox is M k δ(x k, y k, z k ) V k. Thus,

Transcription

1 1 Triple Integrals Mass problem. Find the mass M of a solid whose density (the mass per unit volume) is a continuous nonnegative function δ(x, y, z). 1. Divide the box enclosing into subboxes, and exclude all those subboxes that contain points outside of. Let n be the number of all the subboxes inside, and let V k = x k y k z k be the volume of the k-th subbox. 2. Choose any point (x k, y k, z k ) in the k-th subbox. The mass of the k-th subbox is M k δ(x k, y k, z k ) V k. Thus, M n M k = k=1 n δ(x k, yk, zk) V k = k=1 n δ(x k, yk, zk) x k y k z k k=1 This sum is called the Riemann sum. 3. Take the sides of all the subboxes to 0, and therefore the number of them to infinity, and get n M = lim δ(x n k, yk, zk) V k = δ(x, y, z) dv k=1 The last term is the notation for the limit of the Riemann sum, and it is called the triple integral of δ(x, y, z) over. 1

2 Properties of triple integrals 1. If f, g are continuous on, and c, d are constants, then ( ) c f(x, y, z) + d g(x, y, z) dv = c + d 2. If is divided into two solids 1 and 2, then f(x, y, z) dv g(x, y, z) dv f(x, y, z) dv + f(x, y, z) dv 1 2 The mass of the entire solid is the sum of the masses of the solids 1 and 2. 2

3 Evaluating triple integrals over rectangular boxes a x b, c y d, k z l ˆ b ˆ d ˆ l a c k f(x, y, z) dz dy dx or any permutation, e.g. ˆ d ˆ l ˆ b c k a f(x, y, z) dx dz dy Example. Find the mass of the box if its density is 1 3 x 1 2, 0 y π, 0 z 1 δ(x, y, z) = xz sin(xy) Answer : M = π 1 2π

4 Evaluating triple integrals over simple xy-, xz-, yz-solids A solid is called a simple xy-solid if it is bounded above by a surface z = g 2 (x, y), below by a surface z = g 1 (x, y), and its projection on the xy plane is a region R. [ˆ g2 (x,y) R g 1 (x,y) ] f(x, y, z) dz da Example. Find the mass of the solid defined by the inequalities π 6 y π 2, y x π 2, 0 z xy if its density is δ(x, y, z) = cos(z/y) Answer : M = 5π

5 Volume of = dv Example. Find V of bounded by the surfaces Answer : V = y = x 2, y + z = 4, z = 0 Integration in other orders A simple xz-solid [ˆ g2 (x,z) R g 1 (x,z) ] f(x, y, z) dy da A simple yz-solid [ˆ g2 (y,z) R g 1 (y,z) ] f(x, y, z) dx da 5

6 2 Centre of gravity and centroid Mass, centre of gravity and centroid of a lamina Recall that a lamina is an idealised flat object that is thin enough to be viewed as a 2-d plane region R. The mass M of a lamina with density δ(x, y) is M = R δ(x, y) da The centre of gravity of a lamina is a unique point ( x, ȳ) such that the effect of gravity on the lamina is equivalent to that of a single force acting at the point ( x, ȳ) x = 1 M R x δ(x, y) da, ȳ = 1 M R y δ(x, y) da Example. Find the centre of gravity of a lamina with density δ(x, y) = x + 1 bounded by x 2 + (y + 1) 2 = 1 Answer : M = π, x = 1/4, ȳ = 1 6

7 For a homogeneous lamina with δ(x, y) = const, the centre of gravity is called the centroid of the lamina or the centroid of the region R because it does not depend on δ(x, y) = const. x = 1 x da, ȳ = 1 A R A R y da, A = Example. Find the centroid of a region bounded by (x 1) 2 + y 2 = 1, and (x 2) 2 + y 2 = 4 Answer : A = 3π, x = 7/3, ȳ = 0 R da 7

8 Mass, centre of gravity and centroid of a solid The centre of gravity of a solid with density δ(x, y, z) is a unique point ( x, ȳ, z) such that the effect of gravity on the solid is equivalent to that of a single force acting at the point ( x, ȳ, z) x = 1 M x δ(x, y, z) dv, ȳ = 1 M y δ(x, y, z) dv z = 1 M z δ(x, y, z) dv, M = δ(x, y, z) dv Example. Find the centre of gravity of a solid bounded by the surfaces x 2 + y 2 = 1 and x 2 + y 2 = 4, above by the surface z = 5 x 2 y 2, and below by the surface z = 0 with the density δ(x, y, z) = e 5 x2 y 2 z Answer : M = π(e 4 e 3) , z = 2e4 2e 21 2(e 4 e 3) 0.85 For a homogeneous solid with δ(x, y) = const, the centre of gravity is called the centroid of the solid. x = 1 V z = 1 V x dv, ȳ = 1 V z dv, V = dv y dv Example. Find the centroid of the solid below z = 10 x 2 y 2, inside of x 2 + y 2 = 1, and above z = 0 Answer : V = 19π/2, x = 0, ȳ = 0, z = 271/

9 3 Triple integrals in cylindrical and spherical coordinates Cylindrical coordinates Cylindrical wedge or cylindrical element of volume is interior of intersection of two cylinders: r = r 1, r = r 2 two half-planes: θ = θ 1, θ = θ 2 two planes: z = z 1, z = z 2 The dimensions: θ 2 θ 1, r 2 r 1, z 2 z 1 thickness and height of the wedge. are called the central angle, Divide by cylindrical wedges f(r, θ, z) dv = lim n f(rk, θk, zk) V k n=1 V k = [ area of base ] [ height ] = r k r k θ k z k 9

10 Theorem. Let be a solid whose upper suface is z = g 2 (r, θ) and whose lower suface is z = g 1 (r, θ) in cylindrical coordinates. If projection of on the xy-plane is a simple polar region R, and if f(r, θ, z) is continuous on, then f(r, θ, z) dv = = [ˆ g2 (r,θ) R g 1 (r,θ) ˆ θ2 ˆ r2 (θ) ˆ g2 (r,θ) θ 1 r 1 (θ) g 1 (r,θ) f(r, θ, z) dz ] da f(r, θ, z) r dz dr dθ Example. V and centroid of bounded above by z = 25 x 2 y 2, below by z = 0, and laterally by x 2 + y 2 = 9. Answer : V = 122π/3, z = 1107/488 Converting triple integrals from rectangular to cylindrical coordinates f(r cos θ, r sin θ, z) r dz dr dθ Example. appropriate limits ˆ 3 ˆ 9 x 2 ˆ 9 x 2 y x 2 0 x 2 dz dy dx = π 10

11 Spherical coordinates Spherical wedge or spherical element of volume is interior of intersection of two spheres: ρ = ρ 1, ρ = ρ 2 two half-planes: θ = θ 1, θ = θ 2 nappes of two right circular cones: φ = φ 1, φ = φ 2 The numbers: θ 2 θ 1, ρ 2 ρ 1, φ 2 φ 1 are the dimensions of the wedge. Divide by spherical wedges f(r, θ, φ) dv = lim n f(rk, θk, φ k) V k n=1 V k = (ρ k) 2 sin φ k ρ k φ k θ k f(r, θ, φ) dv = appropriate limits f(r, θ, φ) ρ 2 sin φ dρ dφ dθ 11

12 Example. V and centroid of bounded above by x 2 + y 2 + z 2 = 16 and below by z = x 2 + y 2. Answer : V = 64(2 2)π/3 > 0, z = 3/2(2 2) 2.56 Converting triple integrals from rectangular to spherical coordinates appropriate limits f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ 2 sin φ dρ dφ dθ Example. ˆ 2 ˆ 4 x 2 ˆ 4 x 2 y x 2 0 z 2 x 2 + y 2 + z 2 dz dy dx = 64 9 π 12