Double Integrals, Iterated Integrals, Cross-sections

Transcription

1 Chapter 14 Multiple Integrals 1 ouble Integrals, Iterated Integrals, Cross-sections 2 ouble Integrals over more general regions, efinition, Evaluation of ouble Integrals, Properties of ouble Integrals 3 Area and Volume by ouble Integration, Volume by Iterated Integrals, Volume between Two surfaces 4 ouble Integrals in Polar Coordinates, More general Regions 5 Applications of ouble Integrals, Volume and First Theorem of Pappus, Surface Area and Second Theorem of Pappus, Moments of Inertia 6 Triple Integrals, Iterated Triple Integrals 7 Integration in Cylindrical and Spherical Coordinates 8 Surface Area, Surface Area of Parametric Surfaces, Surfaces Area in Cylindrical Coordinates 9 Change of Variables in Multiple Integrals, Jacobian

2 Chapter 14 Multiple Integrals 146 Average value of a function 147 Cylindrical coordinates, and Spherical Coordinates 147 Integration in cylindrical coordinates, and spherical coordinates 147 Mass, Moments, Centroid, Moment of Inertia

3 Applications of Triple Integrals Suppose an object, in region in R 3, is made of different material in which the density (mass per unit volume) is given by δ(x, y, z), depending on the location (x, y, z) Then the total mass of is given approximately by the Riemann sum i δ(x i, y i, z i ) V i, which will converges to the triple integral m = δ(x, y, z) dv We call it the mass of the object Similarly, one define the ( center of mass (centroid) of the object by (x, y, z) = ) xδ(x, y, z) dv, yδ(x, y, z) dv, zδ(x, y, z) dv m m m

4 Let be a solid object in R 3, and l be a straight line in R 3 Then the moment of inertia I of around the axis l is p 2 δ(x, y, z) dv, where p = p(x, y, z) is the shortest distance from the point (x, y, z) of to the line l, and δ(x, y, z) is the density of at the point (x, y, z) For the coordinate axii, we have I x = I x-axis = (y 2 + z 2 )δ(x, y, z) dv, I y = I y-axis = (x 2 + z 2 )δ(x, y, z) dv and I z = I z-axis = (x 2 + y 2 )δ(x, y, z)dv For any solid region in R 3, efine the center (ˆx, ŷ, ẑ) of gyration by I ˆx = xm Iy, ŷ = m, ẑ = I z m

5 Cylindrical Coordinates In the cylindrical coordinate system, a point P(x, y, z) in three-dimensional space is represented by the ordered triple (r, θ, z), where r and θ are polar coordinates of the projection Q(x, y, ) of P(x, y, z) onto the xy-plane and z is the directed distance from the xy-plane to P To convert from cylindrical to rectangular coordinates, we use the equations x = r cos θ, y = r sin θ, z = z To convert from rectangular to cylindrical coordinates, we use the equations r = x 2 + y 2, tan θ = y x, z = z

6 z 2 = r 2 r = c (c > ) Remark Recall in polar coordinates, r 2 = x 2 + y 2 Hence, the equation z 2 = r 2 can be expressed as z 2 = x 2 + y 2, which is a circular cone Another way to understand is to look 1 = z2 is a cone x 2 = PQ 2 OQ 2 = tan 2 POQ, ie, POQ = π/4, which

7 Theorem If is a solid described in the cylindrical coordinates as { ( r, θ, z ) α θ β, r 1 (θ) r r 2 (θ), z min (r, θ) z z max (r, θ) }, and f (x, y, z) is a continuous function defined in, β r2 (θ) zmax (r, θ) then f (x, y, z) dv = f (r cos θ, r sin θ, z) r dz dr dθ α r 1 (θ) z min (r, θ)

8 efinition The average value of a function f (x, y, z) defined on a region in R 3, to be [f ] average = f dv dv Example Solid = { (x, y, z) x 2 + y 2 2, y, z 3 } is in the shape of a half-cylinder of radius 2 and height 3 The temperature (in C) at (x, y, z) is given by T(x, y, z) = 2yz 2 x 2 + y 2 Find the average temperature in the tank Solution We describe the tank in cylindrical coordinates as { (r, θ, z ) θ π, r 2, z 3 } T(x,y,z) dv Recall the formula [T] average = dv Then 2 π 3 T dv = 2(r sin θ) z 2 r r dz dθ dr ( 2 ) ( π ) ( 3 ) = 2 r 3 dr sin θ dθ z 2 dz = = 144 Similarly, 2 π 3 ( 2 ) ( π ) ( 3 ) dv = r dz dθ dr = r dr dθ dz = 2 π 3 = 6π Hence, the average temperature in the tank is 144 6π = 24 π

9 Example Evaluate x 2 4 x 2 2 x2+y2 (x2 + y 2 ) dzdydx Solution Let be the domain of integral in rectangular coordinates, and R be its shadow in xy-plane Rewrite the iterated integral 2 4 x (x2 + y 2 ) dzdydx = (x2 + y 2 ) dz da yx x2+y2 x2+y2 2 4 x 2 y min (x) y max (x) Then R = { (x, y, ) 2 x 2, 4 x 2 y 4 x 2 }, with boundary curves are y min (x) = 4 x 2, and y max (x) = 4 x 2 } Squaring both, we get a single equation y 2 (x) = (± 4 x 2 ) 2 = 4 x 2, ie y 2 + x 2 = 4, which represents a complete circle (as 2 x 2) R

10 2 4 x 2 2 Example Evaluate 2 (x2 + y 2 ) dzdydx 4 x 2 x2+y2 Solution The domain of integral in rectangular coordinates, and R be its shadow in xy-plane The points of intersection of the (top) plane Then in cylindrical coordinates, can be described as { (r, θ, z) r 2, θ 2π, r z 2 } Then by means of Fubini Theorem, and formula of change of variables, we can simplify iterated integral 2 4 x 2 2 (x2 + y 2 ) dzdydx = (x 2 + y 2 ) dv x2+y2 2 4 x 2 2π = r 2 r dz dr dθ = 2π (2 r)r 3 dr r [ 2r 4 ] 2 = 2π 4 r5 = 26 π = 16π 5

11 Example A solid T is bounded by the cone z = x 2 + y 2 and the plane z = 2 The mass density at any point of the solid T is proportional to the distance between the axis of the cone and the point Find the mass of T Solution Since the density of the solid at a point P(x, y, z) is proportional to the distance from the z-axis to the point P(x, y, z), we see that the density function δ(x, y, z) = k x 2 + y 2 for some constant k Moreover, T can be described as { (x, y, z) x 2 + y 2 4, x 2 + y 2 z 2 } In terms of cylindrical coordinates, T may be given as { (r, θ, z) r 2, θ 2π, r z 2 } So the mass of the solid T is 2π 2 2 δ(x, y, z) dv = k x 2 + y 2 dv = k r r dz dr dθ = 2kπ T 2 [ 2r (2 r)r 2 3 dr = 2kπ 3 r4 4 T ] 2 = 2kπ( 16 3 r 4) = 8kπ 3

12 Example Find the volume of the solid region bounded below by the paraboloid z = x 2 + y 2, and above by the plane z = 2x Solution First we sketch the graphs of the paraboloid S 1 and the plane S 2, and we want to find the common intersection of S 1 and S 2 Let P(x, y, z) be any point of the common intersection, ie they satisfy both 2x = z = x 2 + y 2, so (x 1) 2 + y 2 = 1, which represents a cylinder in space In fact, the intersection is a curve obtained by the intersection of the plane and the cylinder above, which is a bounded closed curve Inside the cylinder, we have (x 1) 2 + y 2 1, ie x 2 + y 2 2x, or equivalently the graph of S 1 is below the graph of the plane S 2 Hence the solid region = { (x, y, z) x 2 + y 2 2x, and x 2 + y 2 z 2x } Then one can switch to cylindrical coordinates to describe as { (r, θ, z) r 2 cos θ, π 2 θ π 2 and r2 z 2r cos θ } The ( 2x ) volume of = 1 dv = 1 dz da = x 2 +y 2 2x x 2 +y 2 π/2 2 cos θ ( 2r cos θ ) 1 dz r drdθ π/2 r 2

13 Example Find the volume of the region that lies inside both the sphere x 2 + y 2 + z 2 = 4 and the cylinder x 2 + y 2 2x = Solution The cylinder S is given by (x 1) 2 + y 2 = 1, so any point P(x, y, z) inside S is given by (x 1) 2 + y 2 1 In terms of cylindrical coordinates, it can be described by (r, θ, z) by r 2 cos 2 θ + r 2 sin 2 θ 2r cos θ, ie r 2 cos θ, it follows that the region can be described in terms of cylindrical coordinates as { (r, θ, z) π 2 θ π 2, r 2 cos θ and 4 r 2 z 4 r 2 } It follows from the definition of triple integral ( that the volume of the region π/2 2 cos θ ) 4 r 2 = 1 dv = r drdθ = π/2 4 r 2 dz π/2 2 cos θ π/2 2 4 r 2 rdrdθ = 2 π/2 π/2 [ (4 r 2 ) 3/2 3 ] 2 cos θ dθ =

14 Remark In the calculation above, one needs to pay attention to the following : (sin 2 θ) 3/2 = sin 3 θ = sin 3 θ for all θ [ π 2, π 2 ] [ ] 2 cos θ π/2 (4 r 2 ) 3/2 dθ π/2 3 [ ] π/2 (4 4 cos = 2 θ) 3/2 (4 4 cos 2 ) 3/2 dθ 3 = = 1 3 π/2 π/2 = 2 3 = 16 3 π/2 π/2 π/2 π/2 π/2 (4 sin 2 θ) 3/2 dθ 3 2 sin θ 3 dθ 2 sin θ 3 dθ sin 3 θ dθ

15 Example Find the volume and the centroid of solid, with same density, bounded by paraboloid S : z = b(x 2 + y 2 ) (b > ) and plane π : z = h (h > ) Solution The intersection of plane π : z = h and the paraboloid S : z = b(x 2 + y 2 ) is { (x, y, h) x 2 + y 2 = ( h/b ) 2 } is described as { (x, y, z) x 2 + y 2 h b, b(x2 + y 2 ) z b }, in polar h coordinates as { (r, θ, z) θ 2π, r b, br 2 z b } 2π h b h The volume of the solid is given by dv = r dz dr dθ br2 h ( b 1 = 2π (hr br 3 ) dz dr = 2π 2 ha2 1 ) 4 ba4 = πh2 2b It follows from rotational symmetry of the region about z-axis, and constant density δ = 1, (x, y) = (, ) Then z = 1 z dv = 2b 2π h b h h ( 4b b h V πh 2 rz dz dr dθ = 2 r br2 h 2 2 b2 r 5 ) dr ( 2 = 4b h 2 h h 2 ( b )2 b2 2 1 ) h 6 ( b )6 = 2 3 h

16 Spherical Coordinates ρ = x 2 + y 2 + z 2 = OP = length of vector OP ϕ = angle between the vector OP and the z-axis, ϕ π θ = angle between the shadow of vector OP onto xy-plane and the x- axis, θ 2π x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ

17

18 Ẹxample Find a spherical coordinate equation for the cone z = x 2 + y 2 Solution Substitute for x, y, and z with spherical coordinates: ρ cos ϕ = z = x 2 + y 2 = (ρ sin ϕ cos θ) 2 + (ρ sin ϕ sin θ) 2 = ρ sin ϕ ie cos ϕ = sin ϕ, so tan ϕ = 1 As ϕ π, we have ϕ = π/4 Example Find a spherical coordinate equation for the sphere x 2 + y 2 + (z a) 2 = a 2 Solution Substitute for x, y, and z with spherical coordinates: a 2 = x 2 + y 2 + (z a) 2 = (ρ sin ϕ cos θ) 2 + (ρ sin ϕ sin θ) 2 + (ρ cos ϕ a) 2 = (ρ sin ϕ) 2 + (ρ cos ϕ) 2 2aρ cos ϕ + a 2 = ρ 2 2aρ cos ϕ + a 2, ie ρ 2 = 2aρ cos ϕ, hence ρ = 2a cos ϕ Remark One should note that x 2 + y 2 = ρ 2 sin 2 ϕ Example Rewrite the sphere x 2 + y 2 + z 2 = 2az (a > ) in spherical coordinates Solution ρ 2 = x 2 + y 2 + z 2 = 2az = 2aρ cos ϕ, so ρ = 2a cos ϕ ( ϕ π)

19 Theorem If is a solid described in the spherical coordinates as { (ρ, ϕ, θ) α θ β, ϕ 1 (θ) ϕ ϕ 2 (θ), ρ 1 (ϕ, θ) ρ ρ 2 (ϕ, θ) }, and f (x, y, z) is a continuous function defined in, then f (x, y, z) dv = β ϕ2 (θ) ρ2 (ϕ, θ) α ϕ 1 (θ) ρ 1 (ϕ, θ) f (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) ρ 2 sin ϕ dρ dϕ dθ Remark In general, it is hard to see if a solid can be expressed in the form stated above However, solid ball, and cone, or their intersection are typical examples that one can think of

20 = f (x, y, z) dv β ϕ2 (θ) ρ2 (ϕ, θ) α ϕ 1 (θ) ρ 1 (ϕ, θ) f (ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) ρ 2 sin ϕ dρ dϕ dθ 1 Rotate the green door with points (ρ, ϕ, θ) fixed by θ = θ about z-axis from θ min = α to θ max = β to trap the solid 2 In the green plane set by θ, bend the red ray OP with ZOP = ϕ away from z-axis from ϕ min = ϕ 1 (θ) to ϕ max = ϕ 2 (θ), like opening a Chinese paper fan 3 Extend the red ray in polar coordinate (ρ, ϕ, θ) with fixed ϕ and θ from ρ from ρ min = ρ 1 (ϕ, θ) and ρ max = ρ 2 (ϕ, θ)

21 Example Prove that the volume of a solid sphere of radius a is 4 3 πa3 Solution Let be described in spherical coordinates as { (ρ, ϕ, θ) θ 2π, ϕ π, ρ a }, then the volume of the solid 2π π a sphere is 1 dv = ρ 2 sin ϕ dρ dϕ dθ π [ ρ 3 ] a = 2π sin ϕ dϕ = 2π a3 π 3 3 sin ϕ dϕ = 4 3 πa3

22 Example Evaluate e (x2 +y 2 +z 2 ) 3/2 dv, where = { (x, y, z) x 2 + y 2 + z 2 a } is the solid ball of radius a Solution We use spherical coordinates to describe the solid ball as { (ρ, ϕ, θ) θ 2π, ϕ π, ρ a } in spherical coordinate system, then e (x2 +y 2 +z 2 ) 3/2 = e (ρ2 ) 3/2 = e ρ3, 2π π a and e (x2 +y 2 +z 2 ) 3/2 dv = e ρ3 ρ 2 sin ϕ dρ dϕ dθ [ π ] [ a ] [ ] a = 2π sin ϕ dϕ e ρ3 ρ 2 e dρ = 2π[1 ( 1)] ρ3 = 4π 3 3 (ea3 1)

23 Example A solid is cut from a solid ball of radius 1 centered at the origin by the inequalities z and y x The mass density of at a point (x, y, z) is given by the function δ(x, y, z) = 3z 2 kg /m 3 (a) Find the total mass of S (b) Find the average mass density of S Solution (a) is described in spherical coordinates by { (ρ, θ, ϕ) ϕ π/2, π/4 θ 5π/4 ρ 1 } The total mass of is π/2 5π/4 1 δ(x, y, z) dv = 3(ρ cos ϕ) 2 ρ 2 sin ϕ dρ dθ dϕ ( π/4 π/2 ) ( 5π/4 ) ( 1 ) = 3 cos 2 ϕ sin ϕ dϕ dθ ρ 3 dρ = 2π π/4 π/2 5π/4 1 (b) The volume of is dv = ρ 2 sin ϕ dρ dθ dϕ ( π/4 π/2 ) ( 5π/4 ) ( 1 ) = sin ϕ dϕ dθ ρ 2 dρ = 1 π 1 3 = π 3 π/4 The average mass of is 2π π/3 = 6

24 Example Use spherical coordinates to find the volume of the solid that lies above the cone z = x 2 + y 2 and below the sphere x 2 + y 2 + z 2 = z Solution The sphere x 2 + y 2 + z 2 = z can be written as ρ 2 = ρ cos ϕ, ie ρ = cos ϕ Since the upper solid cone bounded by z = x 2 + y 2 can be best described in spherical coordinate by ϕ π 4 can be parameterized by { (ρ, ϕ, θ) θ 2π, ϕ π/4, ρ cos ϕ } in terms of spherical coordinates For any point P(x, y, z) in, the distance from the point P to the xy-plane is z = ρ cos ϕ The volume of is 2π π/4 cos ϕ dv = ρ 2 sin ϕ dρ dϕ dθ = = 5π 9 2

25 Example Let be a region bounded above by the sphere x 2 + y 2 + z 2 = 3 2 and below by the cone z = x 2 + y 2 The mass density at any point in is its distance from the xy-plane Find the total mass m of Solution Since the cone z = x 2 + y 2 can be best described in spherical coordinate by ρ π 4 In terms of spherical coordinates, can be parameterized by { (ρ, ϕ, θ) θ 2π, ϕ π/4, ρ 3 } For any point P(x, y, z) in, the distance from the point P to the xy-plane is z = ρ cos ϕ 2π π/4 3 The total mass m = zdv = ρ cos ϕ ρ 2 sin ϕ dρdϕdθ = 2π 3 ρ 3 dρ π/4 sin ϕ cos ϕdϕ = 2π 34 4 sin2 (π/4) = 81π 2 8

26 Example Let be ice-cream solid bounded above by the sphere x 2 + y 2 + z 2 = 2az and below by the cone z = 3(x 2 + y 2 ) Find the volume of Solution The sphere x 2 + y 2 + z 2 = 2az is written in spherical coordinates as ρ 2 = x 2 + y 2 + z 2 = 2az = 2aρ cos ϕ, ie ρ = 2a cos ϕ The cone z = 3(x 2 + y 2 ) is written in spherical coordinates as ρ cos ϕ = 3ρ 2 sin 2 ϕ, ie tan ϕ = 1 and so ϕ = π/6 3 In terms of spherical coordinates, can be described as { (ρ, ϕ, θ) θ 2π, ϕ π/6, ρ 2a cos ϕ } The volume of is given by 1 dv = 2π π/6 2a cos ϕ = 2π 8π 3 π/6 ρ 2 sin ϕ dρdϕdθ a 3 cos 3 ϕ sin ϕ dϕ = 16πa3 3 [ 1 ] π/6 4 cos4 ϕ = 7 12 πa3