Figure 1: u2 (x,y) D u 1 (x,y) yρ(x, y, z)dv. xρ(x, y, z)dv. M yz = E


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1 Contiune on.7 Triple Integrals Figure 1: [ ] u2 (x,y) f(x, y, z)dv = f(x, y, z)dz da u 1 (x,y) Applications of Triple Integrals Let be a solid region with a density function ρ(x, y, z). Volume: V () = 1dV Mass: m = ρ(x, y, z)dv Moments about the coordinate planes: M xy = zρ(x, y, z)dv Center of mass: ( x, ȳ, z) M xz = M yz = yρ(x, y, z)dv xρ(x, y, z)dv x = M yz /m, ȳ = M xz /m, z = M xy /m. Remark: The center of mass is just the weighted average of the coordinate functions over the solid region. If ρ(x, y, z) = 1, the mass of the solid equals its volume and the center of mass is also called the centroid of the solid. xample Find the volume of the solid region between y = 4 x 2 z 2 and y = x 2 +z 2. 1
2 Soln: is described by x 2 + z 2 y 4 x 2 z 2 over a disk in the xzplane whose radius is given by the intersection of the two surfaces: y = 4 x 2 z 2 and y = x 2 + z 2. 4 x 2 z 2 = x 2 + z 2 x 2 + z 2 = 2. So the radius is 2. Therefore [ ] 4 x 2 z 2 V () = 1dV = 1dy da = 4 2(x 2 + z 2 )da = x 2 +z 2 2π 2 (4 2r 2 )rdrdθ = 2π [2r 2 12 ] 2 r4 = 4π xample Find the mass of the solid region bounded by the sheet z = 1 x 2 and the planes z =, y =, y = 1 with a density function ρ(x, y, z) = z(y + 2). Figure 2: Soln: The top surface of the solid is z = 1 x 2 and the bottom surface is z = over the region in the xyplane which is bounded by the other equations in the xyplane and the intersection of the top and bottom surfaces. The intersection gives 1 x 2 = x = ±1. Therefore is a square [, 1] [, 1]. [ ] 1 x 2 m = ρ(x, y, z)dv = z(y + 2)dV = z(y + 2)dz da = x 2 z(y + 2)dzdxdy = (1 x 2 ) 2 (y + 2)dxdy = (y + 2)dy = 32/15 xample Find the centroid of the solid above the paraboloid z = x 2 + y 2 and below the plane z = 4. Soln: The top surface of the solid is z = 4 and the bottom surface is z = x 2 + y 2 over the region defined in the xyplane by the intersection of the top and bottom surfaces. 2
3 Figure 3: The intersection gives 4 = x 2 + y 2. Therefore is a disk of radius 2. By the symmetry principle, x = ȳ =. We only compute z: [ 4 ] 2π m = 1dV = 1dz da = 4 (x 2 + y 2 )da = x 2 +y 2 M xy = zdv = 2π 2 [ 4 ] zdz da = x 2 +y 2 (8 1 2 r4 )rdrdθ = 2π Therefore z = M xy /m = 8/3 and the centroid is (,, 8/3) (x2 + y 2 ) 2 da = [4r r6 ] 2 dθ = 64π/3..8 Triple Integrals in Cylindrical and Spherical Coordinates (4 r 2 )rdrdθ = 8π 1. Triple Integrals in Cylindrical Coordinates A point in space can be located by using polar coordinates r, θ in the xyplane and z in the vertical direction. Some equations in cylindrical coordinates (plug in x = r cos(θ), y = r sin(θ)): Cylinder: x 2 + y 2 = a 2 r 2 = a 2 r = a; Sphere: x 2 + y 2 + z 2 = a 2 r 2 + z 2 = a 2 ; Cone: z 2 = a 2 (x 2 + y 2 ) z = ar; Paraboloid: z = a(x 2 + y 2 ) z = ar 2. The formula for triple integration in cylindrical coordinates: If a solid is the region between z = u 2 (x, y) and z = u 1 (x, y) over a domain in the xyplane, which is described in polar coordinates by α θ β, h 1 (θ) r h 2 (θ), we plug 3
4 Figure 4: in x = r cos(θ), y = r sin(θ) Note: dv rdzdrdθ f(x, y, z)dv = u 1 (x,y) β h2 (θ) u2 (r cos θ,r sinθ) α h 1 (θ) u 1 (r cos θ,r sin θ) [ ] u2 (x,y) f(x, y, z)dz da = f(r cosθ, r sin θ, z)rdzdrdθ xample valuate zdv where is the portion of the solid sphere x2 +y 2 +z 2 9 that is inside the cylinder x 2 + y 2 = 1 and above the cone x 2 + y 2 = z 2. Figure 5: Soln: The top surface is z = u 2 (x, y) = 9 x 2 y 2 = 9 r 2 and the bottom surface is z = u 1 (x, y) = x 2 + y 2 = r over the region defined by the intersection of the top (or 4
5 bottom) and the cylinder which is a disk x 2 + y 2 1 or r 1 in the xyplane. [ ] 9 r 2 2π 1 9 r 2 zdv = zdz da = zrdzdrdθ = 2π [9 2r2 ]rdrdθ = r 2π [9r 2r3 ]drdθ = r 2π [9/4 1/4]dθ = 4π xample Find the volume of the portion of the sphere x 2 +y 2 +z 2 = 4 inside the cylinder (y 1) 2 + x 2 = 1. Figure 6: Soln: The top surface is z = 4 x 2 y 2 = 4 r 2 and the bottom is z = 4 x 2 y 2 = 4 r 2 over the region defined by the cylinder equation in the xyplane. So rewrite the cylinder equation x 2 + (y 1) 2 = 1 as x 2 + y 2 2y + 1 = 1 r 2 = 2r sin(θ) r = 2 sin(θ). V () = 1dV = 2sin(θ) 4 r 2 4 r 2 1dzdA = 2sin(θ) 4 r 2 4 r 2 1rdzdrdθ = 2r 4 r 2 drdθ (by substitution u = 4 r 2 ) = 2 3 [(4 4 sin2 (θ)) 3/2 (4) 3/2 ]dθ (use identity 1 = cos 2 (θ) + sin 2 (θ)) = 3 [1 cos(θ) 3 ]dθ = /2 /2 3 [1 cos3 (θ)]dθ + 3 [1 (1 sin2 θ) cosθ]dθ + π/2 π/2 3 [1 + cos3 (θ)]dθ = 3 [1 + (1 sin2 θ) cosθ]dθ = /3[(θ sin θ + sin 3 θ/3) π/2 + (θ + sin θ sin 3 θ/3) π π/2] = π/3 64/9 2. Triple Integrals in Spherical Coordinates 5
6 Figure 7: In spherical coordinates, a point is located in space by longitude, latitude, and radial distance. Longitude: θ 2π; Latitude: φ π; Radial distance: ρ = x 2 + y 2 + z 2. From r = ρ sin(φ) x = r cos(θ) = ρ sin(φ) cos(θ) y = r sin(θ) = ρ sin(φ) sin(θ) z = ρ cos(φ) Some equations in spherical coordinates: Sphere: x 2 + y 2 + z 2 = a 2 ρ = a Cone: z 2 = a 2 (x 2 + y 2 ) cos 2 (φ) = a 2 sin 2 (φ) Cylinder: x 2 + y 2 = a 2 r = a or ρ sin(φ) = a Figure 8: Spherical wedge element 6
7 The volume element in spherical coordinates is a spherical wedge with sides dρ, ρdφ, rdθ. Replacing r with ρ sin(φ) gives: dv = ρ 2 sin(φ)dρdφdθ For our integrals we are going to restrict down to a spherical wedge. This will mean a ρ b, α θ β, c φ d, f(x, y, z)dv = β d b α c a f(ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ))ρ 2 sin(φ)dρdφdθ Figure 9: One example of the sphere wedge, the lower limit for both ρ and φ are The more general formula for triple integration in spherical coordinates: If a solid is the region between g 1 (θ, φ) ρ g 2 (θ, φ), α θ β, c φ d, then f(x, y, z)dv = β d g2 (θ,φ) α c g 1 (θ,φ) f(ρ sin(φ) cos(θ), ρ sin(φ) sin(θ), ρ cos(φ))ρ 2 sin(φ)dρdφdθ xample Find the volume of the solid region above the cone z 2 = 3(x 2 + y 2 ) (z ) and below the sphere x 2 + y 2 + z 2 = 4. Soln: The sphere x 2 + y 2 + z 2 = 4 in spherical coordinates is ρ = 2. The cone z 2 = 3(x 2 + y 2 ) (z ) in spherical coordinates is z = 3(x 2 + y 2 ) = 3r ρ cos(φ) = 3ρ sin(φ) tan(φ) = 1/ 3 φ = π/6. Thus is defined by ρ 2, φ π/6, θ 2π. V () = 1dV = 2π /6 2π /6 2 ρ 2 sin(φ)dρdφdθ = 3 2 ) 8 π sin(φ)dφdθ = 3 3 (1 7
8 Figure 1: Figure 11: xample Find the centroid of the solid region lying inside the sphere x 2 +y 2 +z 2 = 2z and outside the sphere x 2 + y 2 + z 2 = 1 Soln: By the symmetry principle, the centroid lies on the z axis. Thus we only need to compute z The top surface is x 2 + y 2 + z 2 = 2z ρ 2 = 2ρ cos(φ) or ρ = 2 cos(φ). The bottom surface is x 2 + y 2 + z 2 = 1 ρ = 1. They intersect at 2 cos(φ) = 1 φ = π/3. 2π /3 m = 1dV = 8 3 cos3 (φ) sin(φ)dφdθ 2π /3 2cos(φ) 1 2π /3 ρ 2 sin(φ)dρdφdθ = 1 11π sin(φ)dφdθ =
9 z = M xy /m = 12 zdv = 12 2π 11π 11π [ 12 2π /3 4 cos 5 (φ) sin(φ)dφdθ 11π /3 2cos(φ) 1 2π / π ρ cos(φ)ρ 2 sin(φ)dρdφdθ = ] 1/4 cos(φ) sin(φ)dφdθ [ 4 6 cos6 (φ) π/3 1 4 = sin 2 ] (φ) π/ [9π/8] π xample Convert 3 9 y 2 18 x 2 y 2 x x 2 + y 2 + z 2 dzdxdy into spherical coordinates. 2 +y 2 Soln: We first write down the limits of the variables: y 3 x 9 y 2 x2 + y 2 z 18 x 2 y 2. The range for x tells us that we have a portion of the right half of a disk of radius 3 centered at the origin. Since y, we will have the quarter disk in the first quadrant. Therefore since is in the first quadrant the region,, must be in the first octant and this in turn tells us that we have the following range for θ θ π/2. Now, lets see what the range for z tells us. The lower bound, z = x 2 + y 2 is the upper half of a cone z 2 = x 2 + y 2. The upper bound, z = 18 x 2 y 2 is the upper half of the sphere x 2 + y 2 + z 2 = 18. So the range for ρ ρ 18 Now we try to find the range for φ. We can get it from the equation of the cone. In spherical coordinates, the equation of the cone is 1 = tan(φ), which gives φ = π/4. We have the range for φ φ π/4. Thus 3 9 y 2 18 x 2 y 2 x 2 +y 2 x 2 + y 2 + z 2 dzdxdy = /4 /2 18 ρ 4 sin(φ)dρdθdφ 9
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