Chapter 14 Multiple Integrals 1 Double Integrals, Iterated Integrals, Cross-sections 2 Double Integrals over more general regions, Definition, Evaluation of Double Integrals, Properties of Double Integrals 3 Area and Volume by Double Integration, Volume by Iterated Integrals, Volume between Two surfaces 4 Double Integrals in Polar Coordinates, More general Regions 5 Applications of Double 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 Math 2 in 211
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 Math 2 in 211
Applications of Triple Integrals Suppose an object, in region D 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 D 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) = ) 1 1 1 xδ(x, y, z) dv, yδ(x, y, z) dv, zδ(x, y, z) dv m D m D m D D Math 2 in 211
Let D be a solid object in R 3, and l be a straight line in R 3 Then the moment of inertia I of D 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 D to the line l, and δ(x, y, z) is the density of D 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, D I y = I y-axis = (x 2 + z 2 )δ(x, y, z) dv and D I z = I z-axis = (x 2 + y 2 )δ(x, y, z)dv D For any solid region D in R 3, Define the center (ˆx, ŷ, ẑ) of gyration by I ˆx = xm Iy, ŷ = m, ẑ = I z m D Math 2 in 211
Average Value of Function of Several Variables The average value of a function of several variables defined on a region D in R 3, to be [f ] average = D f dv D dv Math 2 in 211
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 Math 2 in 211
z = r r = c (c > ) Math 2 in 211
Theorem If D 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 D, β ϕ2 r2 (θ) then f (x, y, z) dv = f (r cos θ, r sin θ, z) r dz dr dθ D α ϕ 1 r 1 (θ) Math 2 in 211
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 T T r 2 [ 2r = 2kπ (2 r)r 2 3 ] 2 dr = 2kπ 3 r4 = 2kπ( 16 8kπ 4) = 4 3 3 Math 2 in 211
Example Evaluate = = 2 4 x 2 2 4 x 2 2 4 x 2 2 4 x 2 2π 2 2 r 2 2 4 x 2 4 x 2 2 x2+y2 (x2 + y 2 ) dv Solution The domain D of integral in rectangular coordinates, and R be its shadow in xyplane Then in cylindrical coordinates, D can be described as { (r, θ, z) r 2, θ 2π, r z 2 } Then we can simplify integral 2 x2+y2 (x2 + y 2 ) dv 2 (x2 + y 2 ) dv = x2+y2 [ r r 2 4 r dr dθ = 2π 4 ] 2 = 8π D (x 2 + y 2 ) dv Math 2 in 211
Example A tank D is a solid in the shape of a half-cylinder of radius 2 and height 3 It is situated in R 3, given by the inequalities x 2 + y 2 2, y, z 3 The temperature (in C) at the point (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 D in cylindrical coordinates as { (r, θ, z ) θ π, r 2, z 3 } D Recall the formula [T] average = D T dv Then dv 2 π 3 T dv = 2(r sin θ) z 2 r r dz dθ dr D( 2 ) ( π ) ( 3 ) = 2 r 3 dr sin θ dθ z 2 dz = 2 2 4 9 = 144 2 π 3 Similarly, dv = r dz dθ dr ( D 2 ) ( π ) ( 3 ) = r dr dθ dz = 2 π 3 = 6π Hence, the average temperature in the tank is 144 6π = 24 π Math 2 in 211
Example Find the volume of the solid region D 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 D = { (x, y, z) x 2 + y 2 2x, and x 2 + y 2 z 2x } Then one can switch to cylindrical coordinates to describe D as { (r, θ, z) r 2 cos θ, π 2 θ π 2 and r 2 z 2r cos θ } The volume of D = ( 2x ) π/2 1 dz da = x 2 +y 2 2x x 2 +y 2 π/2 D 2 cos θ 1 dv = ( 2r cos θ r 2 ) 1 dz r drdθ Math 2 in 211
Example Find the volume of the region D 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 π/2 2 cos θ ) 4 r 2 region D = 1 dv = r drdθ = D π/2 π/2 2 cos θ π/2 2 4 r 2 rdrdθ = 2 π/2 π/2 4 r 2 dz [ (4 r 2 ) 3/2 Math 2 in 211 3 ] 2 cos θ dθ =
Example Find the volume and the centroid of the solid D, bounded by the paraboloid z = b(x 2 + y 2 ) (b > ) and the plane z = h (h > ) Solution The intersection of the plane z = h and the paraboloid z = b(x 2 + y 2 ) is { (x, y, h) x 2 + y 2 = ( h/b ) 2 } D is described as { (x, y, z) x 2 + y 2 h b, b(x2 + y 2 ) z b }, in polar coordinates h as { (r, θ, z) θ 2π, r b, br 2 z b } The volume of the solid D is given by h b 2π (hr br 3 ) dz dr = 2π D dv = 2π h b h ( 1 2 ha2 1 4 ba4 br ) = πh2 2b 2 r dz dr dθ = Assume that δ(x, y, z) 1 By symmetry, (x, y) = (, ) Then z = 1 z dv V D = 2b 2π h b h h ( 4b b h πh 2 rz dz dr dθ = 2 r br2 h 2 2 b2 r 5 ) dr ( 2 = 4b h 2 h 2 2 1 h 2 ( b )2 b2 2 1 ) h 6 ( b )6 = 2 3 h Math 2 in 211
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 Math 2 in 211
Math 2 in 211
Example 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) Math 2 in 211
Theorem If D is a solid described in the spherical coordinates as { (ρ, ϕ, θ) α θ β, ϕ 1 ϕ ϕ 2, ρ 1 (ϕ, θ) ρ ρ 2 (ϕ, θ) }, and f (x, y, z) is a continuous function defined in D, then f (x, y, z) dv = D β ϕ2 ρ2 (ϕ, θ) α ϕ 1 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 Math 2 in 211
Example Prove that the volume of a solid sphere D of radius a is 4 3 πa 3 Solution Let D be described in spherical coordinates as { (ρ, ϕ, θ) θ 2π, ϕ π, ρ a }, then the volume of the solid sphere D is π [ ρ 3 ] a 2π 3 2π π a sin ϕ dϕ = 2π a3 ρ 2 sin ϕ dρ dϕ dθ = π 3 sin ϕ dϕ = 4 3 πa3 Math 2 in 211
Example Evaluate D e (x2 +y 2 +z 2 ) 3/2 dv, where D = { (x, y, z) x 2 + y 2 + z 2 a } is the solid ball of radius a Solution We use spherical coordinates to describe B as as { (ρ, ϕ, θ) θ 2π, ϕ π, ρ a }, then e (x2 +y 2 +z 2 ) 3/2 = e (ρ2 ) 3/2 = e ρ3, and 2π π a e (x2 +y 2 +z 2 ) 3/2 dv = e ρ3 ρ 2 sin ϕ dρ dϕ dθ = D [ π ] [ a ] [ ] a 2π sin ϕ dϕ e ρ3 ρ 2 e dρ = 2π[1 (1)] ρ3 = 4π 3 3 (ea3 1) Math 2 in 211
Example A solid D is cut from a solid ball of radius 1 centered at the origin by the inequalities z and y x The mass density of D 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) D is described in spherical coordinates by { (ρ, θ, ϕ) ϕ π/2, π/4 θ 5π/4 ρ 1 } The total mass of D is π/2 5π/4 1 δ(x, y, z) dv = 3(ρ cos ϕ) 2 ρ 2 sin ϕ dρ dθ dϕ D( 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 D is dv = ρ 2 sin ϕ dρ dθ dϕ ( D 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 D is 2π π/3 = 6 Math 2 in 211
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 D 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 D, the distance from the point P to the xy-plane is z = ρ cos ϕ The volume of 2π π/4 cos ϕ D is dv = ρ 2 sin ϕ dρ dϕ dθ = 5π D 9 2 Math 2 in 211
Example Let D 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 D is its distance from the xy-plane Find the total mass m of D 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, D can be parameterized by { (ρ, ϕ, θ) θ 2π, ϕ π/4, ρ 3 } For any point P(x, y, z) in D, the distance from the point P to the xy-plane is z = ρ cos ϕ The total mass 2π π/4 3 m = zdv = ρ cos ϕ ρ 2 sin ϕ dρdϕdθ = 2π 3 D ρ 3 dρ π/4 sin ϕ cos ϕdϕ = 2π 34 4 sin2 (π/4) 2 Math 2 in 211 = 81π 8
Example Let D 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 D Solution The sphere x 2 + y 2 + z 2 = 2az is written in spherical coordinates as ρ 2 = 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 D can be described in spherical coordinates as { (ρ, ϕ, θ) θ 2π, ϕ π/6, ρ 2a cos ϕ } The volume of D is given by 2π 8π 3 π/6 cos 3 ϕ sin ϕ dϕ = 16π 3 2π π/6 2a cos ϕ [ 1 4 cos4 ϕ ρ 2 sin ϕ dρdϕdθ = ] π/6 = 7 12 πa3 Math 2 in 211