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Integrating Functions of Several Variables16.1 The Definite Integral of a Function of Two VariablesConsider z = f (x, y) continuous on a bounded region R on the x-y plane. If we divide R into n sub-regionsR1 …Rn of areas ?A1…?Anrespectively, with each sub region Rk containing the point Pk (xk, yk),limn→∞k=1nf(xk,yk)?Ak=Rfx,ydA, where dA can be dxdy or dydx in the Cartesian coordinate system. abfx dx is the area under the curve y = f (x) in [a, b], and if f (x) = 1, abfx dx is the length [a, b]. If we extend this concept one dimension, we can say abcdf x, yd A is the volume under the surfacez = f (x, y) in the rectangle [a, b] x [c, d], and if z = f (x, y) = 1, abcdf x, yd A is the area of the rectangle.Average Value of a FunctionIf f (x, y) is piecewise continuous in a bounded region R with piecewise smooth boundary, then fave=1area Rfx,ydA.eg 1 The table below gives the value of z = f (x, y). R is the rectangle 1 ≤ x ≤ 1.2, 2 ≤ y ≤ 2.4. Find the Riemann sums which are a reasonable under and over-estimates for Rfx,ydA. with Δx = 0.1 and Δy = 0.2. Find the average value of f in R.y/x1.01.11.22.057102.24682.4354For Δx = 0.1 Δy = 0.2, the over-estimate will be the sum of the area of each square times the largest value the function takes at any of the corners any sub-rectangle. Since Δx = 0.1 x Δy = 0.2, the area of each square will be Δx Δy = 0.02. The upper sum will be 0.02(7+6+10+8) = 0.62. The lower sum will be 0.02(4+3+6+4) = 0.34 For Δx = 0.1 Δy = 0.2, we can say that 0.34 < Rfx,ydA < 0.62.We can get a better estimate if we average the two sums, or Rfx,ydA 0.34+0.622=0.48.The average value of the function will be fave=1total area Rfx,ydA=0.480.2(0.4)=6.eg 2 The figure below shows the contours of f (x, y) on a square R. Using Δx = Δy = 12, find Riemann sums which are reasonable over and under-estimate for Rfx,ydA. Repeat the same problem using Δx = Δy = 155388828165For Δx = Δy = 12, the over-estimate will be the sum of the area of each square times the largest value the function takes at any of the corners of that sub-rectangle. If we start counterclockwise with the square at (0, 0), an over-estimate of the integral will be 1412.8+10.8+12+17=13.15.For Δx = Δy = 12, the under-estimate will be the sum of the area of each square times the smallest value the function takes at any of the corners of that sub-rectangle. If we start counterclockwise with the square at (0, 0), an under-estimate of the integral will be 149.8+7+9.8+10.8=9.35.For Δx = Δy = 12, we can say that 9.35 < Rfx,ydA< 13.15.We can get a better estimate if we average the two sums, or Rfx,ydA 13.15+9.352=11.25.If we used the value of the function at the center if the squares, we obtain 1413+11+9.8+10.8=11.15.For Δx = Δy = 1, the over-estimate will be 17 and an under-estimate 7. If we average the two values, we get Rfx,ydA 17+72 = 12.Homework 16.1Approximate the Riemann sum for RMx,ydA in 0≤ x ≤4, 0≤ y ≤4 using four partitions as shown in the figure below.Ans: 312Approximate the Riemann sum for RNx,ydA in 0≤ x ≤40, 0≤ y ≤40 using one partitions as shown in the figure below.Ans: 8480 Table below gives the values of the function Px,y. Find an over and underestimate of RPx,ydA to approximate the integral in 0≤ x ≤6, 0≤ y ≤2 by using four partitionsy/x0360346145725710Ans: Over 87; under 48; estimate 67.5 16.2 Iterated IntegralsIterated Integrals over Rectangular RegionsIf I is an iterated integrals over a rectangular region, the integration can be switched.Fubini’s Theorem:If f (x, y) is continuous in the rectangle R = {(x, y) | a ≤ x ≤ b, c ≤ y ≤ d}, then I=Rfx,ydA=abcdfx,ydydx=cdabfx,ydxdy.eg 3 0102x+y2dydx=01(xy+y33)|y=0y=2dx=01(2x+83)dx=113. 0201 (x+y2)dxdy=02(x22+y2x)|x=0x=1dy=02(12+y2)dy=113. eg 4 The figure below shows the contours of f (x, y) = x2 + y2. Using Δx = Δy = 1, estimateRfx,ydAby finding the average of an over-estimate and under-estimate of the integral for 0 ≤ x, y ≤ 3. Repeat the problem by evaluating the integral.28467310784300Exact will be 0303(x2+y2)dxdy=039+3y2dy=54.The over-estimate will be 1(3 + 6 + 10 + 13 + 9 + 6 + 10 +13 + 18) = 91.The under-estimate will be 1(0 + 2 + 5 + 6 + 3 + 2 + 5 + 6 +9) = 30.The average will be Rfx,ydA≈60.5 with relative error in 12%eg 5 120πxcosxydxdy has to be integrated by parts.Let u = x, du = dx; dv = cos(xy); v = sin(xy)/y.So 120πxcosxydxdy=12(xysinxy)|0π-0πsin(xy)/ydx)dy= 12(πysinπy+1y2cosπy-1y2)dy=12πysinπydy+121y2cosπydy-121y2dy.If we integrate by parts the first integral with u = 1y; du = -1y2;dv=sinπydy; andv=-cosπy/π, the integral becomes-1ycosπy|12-121y2cosπydy+121y2cosπydy+1y|12=-2.This integral will be easier to compute if we switch the integrals.0π12xcosxydydx=0πsin?(xy)|y=1y=2dx=0πsin2x-sin?(x)dx=-2 Iterated Integrals over General RegionsIn general, the limits of integration do not need to be over rectangular regions. Since an integral is a function of its limits, the inside integral should be a function of the outside variable of integration. Type IIf the region of integration in the x-y plan is bounded above by y(x) = g2(x) and below by y(x) = g1(x) in a < x < b, the double integral will be Rfx,ydA=abg1(x)g2(x)f x,ydydxeg 6 Find the area between y = sin(x) and y = cos(x) in 0 ≤ x ≤ π4.558800406400π4sin?(x)cos?(x)dydx=0π4cosx-sinxdx=sinx+cos?(x)|0π4=2-1.eg 7 Evaluate ∫R ex2dA where R is the triangle (0, 0), (1, 0), (1, 1). Since the region is bounded above byy(x) = x and below by y = 0 in 0 < x < 1. The integral becomes010xex2dydx= 01xex2dx=ex22|01=(e-1)2.Type IIIf the region of integration in the x-y plane is bounded at the right by x(y) = h2(y) and at theleft by x(y) = h1(y) in c < y < d, the double integral will be Rfx,ydA=cdh1(y)h2(y)fx,ydxdy eg 8 Consider RydA where R area bounded by y=x,y=2-x and y = 0523875116840A type II integral will be only one. 01y22-yydxdy=012y-y2-y3dy=512 whereas a type I integral will be two double integrals 010xydydx+1202-xydydx. eg 9 Find R 2ydA where R is bounded by y = x2 and y = x. If we use a Type I integral, the region is bounded above by y(x) = x and below by y(x) = x2 in 0 < x < 1. The integral then becomes 01x2x2ydydx=01(y2)|x2xdx=01(x-x4)dx=12-15=310.If we use a Type II integral, the region is bounded at the right by x(y) = y and at the left byx(y) = y2 in 0 < y < 1. The integral then becomes 01y2y2ydxdy=01(2xy)|y2ydy= 01(2y32-2y3)dy=45-12=310.eg 10 Find Rx2dA where R is bounded by xy = 4 and y = x, y = 0, x = 4.523875118110If we use a type I integral, we need two double integrals.020xx2dydx+2404xx2dydx=02x3dx+244xdx=28.If we use a type II integral, we need also two double integrals.01y4x2dxdy+12y4yx2dxdy=28.Reversing the Order of IntegrationSometimes it is easier to reverse the order of integration to make the integration easier.eg 11 To evaluate02x24x1+y2dydx, we need the trig substitution y = tan (θ). The integral then becomes02Arctanx2arctan4[xsec(θ)]dθdx02[xlnsecθ+tanθ|θ=Arctanx2θ=Arctan4dx=02xln17+4-x4+1-x2dx. This type I integral is very complicated. A simpler integral is found if we change the order of integration and make the integral a Type II integral. 1774334372997If we plot the region of integration, we obtain02x24x1+y2dydx=040yx1+y2dxdy=04y21+y2dy=1217-1. eg 12 013y3ex2dxdy cannot be evaluated as it is. If we change the order of integration, by plotting the region of integration, we obtain527685132080030x/3ex2dydx=03xex23dx=(e9-1)6.eg 13 Evaluate ∫R11+x2dA where R is the triangle (0, 0), (0, 1), (1, 1). If we use a type I integral,01x111+x2 dydx=0111+x2-x1+x2dx=tan-11-ln?(2)2.If we use a type II integral, 010y11+x2dxdy=01tan-1ydy. If we then integrate by parts, we obtain01tan-1ydy=ytan-1(y)|01-01y1+y2dy=π4-ln?(2)2.Volumes Between Two SurfacesThe volume between the surface zbottom=g(x,y) and ztop=h(x,y) can be given by∫Rztop-zbottomdA, where dA is either dxdy or dydx. eg 14 Find the volume bounded by the plane x + y + z = 1 in the first octant.Since ztop=1-x-y and zbottom=0, 01x1-x1-x-ydydx=16.eg 15 Find the volume bounded by the circular cylinders x2 + y2 = 1 and x2 + z2 = 1.64015248752 First octant view of the volume VolumeIf we let R be the circle x2 + y2 = 1 with ztop=1-x2 and zbottom=-1-x2, the integral for the volume becomes -11-1-x21-x21-x2—-1-x2dydx=-1121-x2y|-1-x21-x2dx=4-111-x2dx=8011-x2dx=163.eg 16 Find the volume in the first octant of the paraboloid f (x, y) = 2 – x2 – y2.0202-x22-x2-y2dydx=022-x22-x2-2-x233dx=2302(2-x2)3dx.If we apply the trig substitution, 830π2cos4dθ=0π283(1+cos2θ2)2dθ=0π2231+2cos2θ+1+cos4θ2dθ= 2332θ+sin2θ+sin4θ8]|0π2=π2.We can also find volumes of solids by integrating over the projection on the xz plane.If the region R is on the xz plane, the integral becomes ∫Rhx,z-g(x,z)dA , where dA is either dxdz or dzdx. See picture below. Refer to homework problem 13. We can also find volumes of solids by integrating over the projection on the yz plane.If the region R is on the yz plane, the integral becomes ∫Rhy,z-g(y,z)dA , where dA is either dydz or dzdy. See picture below. Refer to homework problem 14. Homework 16.201-11(1+x2y2dxdy Ans: 20/90ln201xexydxdy Ans: 1-ln2ln2 1301ex+ydxdyAns: 01xxx2ydydxAns: -1/40010πxysinxdxdyAns: π/2Find the average value of f(x ,y)=x2y2 in the rectangle 0≤ x ≤1, 0≤ y ≤2.Ans: 4/9Evaluate Rfx,ydydx for f(x ,y)=10x4y where R is the triangle with vertices (0,0),(0,2)(1,1).Ans: 2/3Evaluate Rfx,ydxdy for f(x ,y)=4x3y where R is bounded by y = x2 and y = 2x.Ans: 64/3Evaluate Rfx,ydxdy for f(x ,y)= y2cos(x) where R is bounded by x = y3 and the lines y = 0 and x=π.Ans: -2/3Reverse the order of integration to evaluate 04y4e-x2dxdyAns: 12(1-e-16)Find the volume of the tetrahedron bounded by coordinate planes and the plane with xint =1; yint = 1 and zint =1. Ans: 1/6Find the volume of the solid bounded by the planes x=0, y=0, 2x+2y+z=2and 4x+4y –z = 4. Ans: 1Find the volume of the solid bounded by y= x2+z2 and the plane y =?3 for (x,z) in the rectangle x=0, x=4, z=0,z=5. Ans: 1000/3Find the volume of the solid between the surfaces x= ?z2 and x= z2 +2 for (y,z) in the rectangle y=0, y=4, z=0,z=1. Ans: 32/316.4 Double Integrals in Polar CoordinatesAny point (x, y) in R2 can be expressed in Polar Coordinates as r,θ, where r= x2+y2; tanθ=yx;x=rcosθ;y=rsin(θ) are the transformations equations. The area of the polar function r(θ), with a single integral, is given by Area = 12abr2dθ.eg 17 Find the area of the cardioid r=1+cosθ.Area= 1202π(1+cosθ)2dθ.=1202π(1+2cosθ+cos2θ)dθ.=12θ+2sinθ+θ2+sin2θ4|02π=3π2.The area of the polar function r(θ), with a double integral, is given by Area=Rfx,ydA=θ1θ2r1r2rdrdθ where the element of area dA=rdrdθ.eg 18 Find the area, using double integrals, for the previous problem.Area=02π01+cosθrdrdθ=02π(1+cosθ)22dθ=3π2.eg 19 Find the area outside the circle r = 2 and inside the cardioid r = 2 + 2cosθArea=-π2π222+2cosθrdrdθ=20π2(2+2cosθ)22-222dθ=0π2(8cosθ+4cos2(θ))dθ=.0π2(8cosθ+2(1+cos?(2θ)))dθ=8+πeg 20 Evaluate the volume in the first octant of the paraboloid f (x, y) = 2 – x2 – y2.If we use polar coordinates, V= 0π202(2-r2)rdrdθ=0π2r2-r44|02dθ=0π21dθ=π2.eg 21 Evaluate 0202x-x2x2+y2 dydx.Since there is circular symmetry, we can change the integral to polar coordinates. Since y=2x-x2 orx2 + y2 = 2x, is the semicircle with center (1, 0) and radius 1, r=2cos(θ) in 0 < θ < π2. The integral becomes 0π202cos(θ)rrdrdθ=0π28(cosθ)33dθ=169.eg 22 Find the volume in the first octant between x2 + y2 = 9 and x + z = 3.45899723854In Cartesian Coordinates the integral becomes 0309-x23-xdydx. If we change to polar coordinates, we have 0π203(3-rcosθ)rdrdθ=27π4-9.Homework: 16.4Use polar coordinates to evaluate the ∫Rydxdy where R is the region bonded by the upper half of the cardioid r =1+cos(θ) and the x-axis. Ans: 4/3What is the area inside one leaf of the rose r = 2 cos (3θ)? Ans: π/3Use polar coordinates to evaluate the ∫Rxydxdy where R is the region bonded by the semicircle y=x-x2 and the x-axis. Ans: 1/24Evaluate the ∫Rx2+y2-2dxdy with R = {(x,y): 2 ≤ x2 + y2 ≤ 4}. Ans: π/4Find the value of by first switching it from polar to rectangular coordinates. Ans: The integrals (a) and (b) are given in polar coordinates. Rewrite them as iterated integrals in rectangular coordinates.Ans: , Applications of double integrals: Average Value of a FunctionIf f (x, y) is piecewise continuous in a bounded region R with piecewise smooth boundary, then fave=1area Rfx,ydA=?fx,ydA?dA.eg 23 Find the average value of f (x, y) = 2yey2in 0<x<1;0<y< x.fave=010x2yey2dydx010xdydx=(e-2)23.Centroids (Center of Mass)2269490425450M=imixi020000M=imixiThe moment M (also called first moment) is defined as M = mr where m is the mass and r is moment arm (distance from the particle to the axis). For a discrete system of particles of mass mi: In one dimension, the moment about the origin is given by , so the center of mass will bex=M /imi =imixi/imiIn two dimensions, the moment about the axis is given by Mx=imiyi and My=imixiso the center of mass will be x=Mym=imixi/imiand y=Mxm=imiyi/imiFor a one dimension continuous system with mass density (linear) ρ, the center of mass will bex=Mxm=ρxdxρdx. If the mass density is uniform (constant), x=ρxdxρdx=xdxdx.For a two dimensions thin plate with a uniform (constant) mass (area) density ρ bounded by the functionf (x) – g (x), the center of mass will bex=Mym=abρxfx-gxdxρfx-gxdxand y=Mxm=12abρf2x-g2xdxρfx-gxdx.eg 24 Find the centroid of the triangular lamina bounded by y = 2x, y = 0, x = 1 with uniform density 3gm/cm2.x=Mym=013x2x-0dx0132xdx=23;and y=Mxm=12013[2x)2-02dx0132xdx=23 For a two dimensional thin plate with non-uniform (variable) mass density ρ(x,y).x=Mym=xρx,ydAρdA and y=Mxm=yρx,ydAρdA.eg 25 Find the centroid of the triangular lamina with vertices (0, 0), (1, 0) and (0, 1) with mass density ρx,y=xy kg/m2.x=Mym=0101-xx2y dydx0101-xxy dydx=25 and y=Mxm=0101-xxy2 dydx0101-xxy dydx=25.eg 26 Find the centroid of a lamina shaped in the form of a quarter circle or radius 1 with density proportional to the distance from the center of a circle. Because of the symmetry, x=y. Since ρ=kx2+y2 , if we use polar coordinates, x=y=xρx,ydAρdA=k x2+y2 dAk x2+y2 dA=0π201rcosθrrdrdθ0π201rrdrdθ=0π2cos(θ)4dθ0π213dθ=32π.Moments of Inertia of Plane AreaThe moment of inertia I (also called second moment) is defined as M = mr2 where m is the mass and r is moment arm (distance from the particle to the axis). The moment of inertia of a plane area about an axis relates the angular acceleration about the axis to the force twisting the plane area (torque). The moment of inertia of a plane lamina of density ρ(x,y) about an axis is defined as p2dm where p is the perpendicular distance from a point (x, y) to the axis and dm = ρx,ydA where ρ(x,y) is the area density, dA is the element of area and dm is the element of mass.The moment of inertia about the x-axis will be Ix=y2ρx,ydA.The moment of inertia about the y-axis will be Iy=x2ρx,ydA;The moment of inertia about the origin (polar moment) will be I0=Ix+Iy=(x2+y2)ρx,ydA.eg 27 Find Ix, Iy and I0 of a lamina of constant density k, bounded by x = y2, x = – y2 in – 1 ≤ y ≤ 1.Ix=-11-y2y2y2kdxdy=k-112y4dy=k45; Iy=k-11-y2y2x2dxdy=k-1123y6dy=k421; I0=k104105.eg. Find the moment of inertia about the origin on a lamina bounded by circle x2 + y2 = 4 with density ρx,y=k a constant.I0=Ix+Iy=(x2+y2)ρx,ydA.=02π02r2rdrdθ=8kπ.Area of a Surface Described by z = f (x, y)The area of the surface z = f (x, y), (x, y) in D, where f, fx and fy are continuous is given byAs=∫Dfx2x,y+fy2x,y+1dA.eg 28 Calculate the surface area of the portion of the surface z = xy for D : x2 + y2 ≤ 9.As=∫∫Dy2+x2+1 dA=02π03r2+1 rdrdθ=2π3(r2+1)32=2π3(1032-1).eg 29 Find the surface area of the volume in the first octant bounded by the circular cylindersx2 + y2 = 1 and x2 + z2 = 1.36385526670The area of the volume in the x-y dydx=π4+π4+1+01dx=2+π2 square units.HomeworkFind the average value of f(x ,y)=x2y2 in the rectangle 0≤ x ≤1, 0≤ y ≤2.Ans: 4/9A plate in the xy-plane with distances measured in ft. occupies the region bounded by the parabola y=x ? y2 and the x-axis and its density at (x, y) is 6x2 lb/ft2. (a) How much does the plate weigh? (b) Where is the center of gravity? Ans: (a) 3/10 lb, (b) (2/3,2/21)Find the centroid of the quarter circle lying between y=1-x2 and the y-axis for 0≤ x ≤1 with density ρ=1. Ans: (43π,43π)A flat disk occupies the disk x2 ?2x+ y2 ≤ 0 in the xy –plane with distances measured in inches and, and its area density at (x,y) is ρ=5x2+y2 ounces per square inch. Where is the center of gravity? Ans: (6/5,9/20)16.3 Triple IntegralsLet f (x, y, z) be a continuous function in a regions V in R3. The triple integral over R is Vfx,y,zdV where dV = is an element of volume. If f (x, y, z) = 1, the triple integral will give the volume of the region V.Iterated Integrals over Rectangular RegionsIf I is an iterated integrals over rectangular region, the integration can be switched.I=abcdeffx,y,zdxdydz=cdefabfx,ydydzdx=efabcdfx,ydzdxdy… eg 30 13120π01zxsinxydzdydx=112+3-24π cu units.Iterated Integrals over General RegionsType IIf the region of integration in R3 is bounded above by the function h (x, y) and below by the function g (x, y) in the area A in the xy plane, the triple integral will be Vfx,y,zdV=g(x,y)h(x,y)fx,y,zdzdA, where dA is dxdy or dydx. See figure below. 251963356475eg 31 Find VxdV where R is the region bounded by the plane y = z, and the cylinder x2 + y2 = 1 in the first octant.Since the region is bounded above by y = z and below by the xy plane, the integral becomes0101-y20yx dzdxdy=18.eg 32 Find the volume bounded by the parabolic surface z=y the plane x + y = 1 and the x-y plane.53663513982V=0101-x0y1dzdydx=0101-xy dydx=0123(1-x)32dx=2325(1-x)52|10=415. Type IIIf the region of integration in R3 is bounded in the front by the function g2 (y, z) and in the back by the function g1 (y, z) in the area A in the yz plane, the triple integral will be Vfx,y,zdV=g(y,z)h(y,z)fx,y,zdxdA , where d A is dydz or dzdy. See figure below. eg. 33 Set VxdV where V is the region bounded by the plane x = 1, and the paraboloid x = y2 + z2.70040545720Since the region is bounded in the back by x = y2 + z2 and in the front by the x = 1 plane, the integral becomes-11-1-z21-z2y2+z21xdxdydz.A simple way of solving this integral is defining a polar coordinate system in the y-z plane such thaty=rcosθ, z=rsinθ with r2=x2+y2 and tanθ=zy .If we do so, the integral becomes 02π01r21x dxrdrdθ=π3 .Type IIIIf the region of integration in R3 is bounded in the right by the function g2 (x, z) and in the left by the function g1 (x, z) in the area A in the yz plane, the triple integral will beVfx,y,zdV=g(x,z)h(x,z)fx,y,zdydA , where d A is dxdz or dzdxeg 34 Find VydV where V is the region bounded by the cylinder x2 + z2 = 4, and the planes y = 0 and760730133985y = 6.Since the region is bounded in the right by the plane y = 0 and in the right by the y = 6 plane, the integral becomes -22-4-x24-x206y dydzdx=72π. A simple way of solving this integral is defining a polar coordinate system in the x-z plane such that x=rcosθ, z=rsinθ with r2=x2+z2 and tanθ=zx .If we do so, the integral becomes 02π0206ydyrdrdθ=72π.Average Value of a FunctionIf f (x, y, z) is piecewise continuous is a bounded region with piecewise smooth boundary, thenfave=1volumefx,y,zdV=fx,y,zdV ∫∫∫dV.eg 35 Find the average value of f (x, y, z) = xyz in 0 < x < 1; 0 < y < x3; 0 < z < 8.As a Type I integral, fave=010x308xyz dzdydx010x308dzdydx=22=1 CentroidsDefinition: If ρx,y,z is a mass (weight) density, the mass (weight) is given bym=∫∫∫v ρ(x,y,z) dV .The centroid will be x=Myzm=∫∫xρ x,y,zdV∫∫ρdV, y=Mxzm=∫∫yρ x,y,zdV∫∫ρdV, z=Mxym=∫∫zρ x,y,zdV∫∫ρdV Moments of Inertia of a Solid BodyThe moment of inertia of a solid body about an axis relates the angular acceleration about the axis to the force twisting the solid (torque). Moment of inertia of a solid body of density ρ(x,y) about an axis is defined asp2dm, where p is the perpendicular distance from a point (x,y,z) to the axis and dm = ρx,y,zdV where ρx,y,z is the volume density, dV is the element of volume and dm is the element of mass.The moment of inertia about the x-axis will be Ix=(y2+z2) ρ x,y,zdV.The moment of inertia about the y-axis will be Iy=(x2+z2) ρ x,y,zdV.The moment of inertia about the y-axis will be Iz=(x2+y2) ρ x,y,zdV.Homework: 16.3Express Vfx,y,zdV as an integral where V is the solid bounded by z = 0, z =1?y2, x = ?2 and x = 0 as a Type I integral.Ans -20-1101-y2fdzdydx What is the value of if F is the constant function F(x, y, z) = 7 and V is a bounded solid with piecewise smooth boundary whose volume is 10? (b) What is the average value of F in V for the function and solid part (a)? Ans: 70,7A block occupies the region bounded by x = –2, x = 2, y = –2, y = 2, z = 1, and z = 2 in xyz-space with distances measured in meters and its density at (x, y, z) is x2ey sin z kilograms per cubic meter. What is its mass? Ans: kg.A solid V bounded by z = x2 + 1, z = –y2 – 1, x = 0, x = 2, y = –1, and y = 1 in xyz-space with distances measured in feet contains electrical charges with density 8xy2 coulombs per cubic foot at (x, y, z). What is the overall net charge in V ? Ans:736/15What is the average value of f = xy3z7 in the box V bounded by x = 0, x = 2, y = 0, y = 2, z = 0 and z = 2? Ans:32Describe the solids of integration that lead to the iterated integrals: Ans: one fourth of an upper hemisphere centered at the origin with radius 2, with x ≥0 and y ≥ 0.Describe the solids of integration that lead to the iterated integrals:Ans: The tetrahedron bounded by the plane x-4y-2z=4 and the coordinate planes.Evaluate with V bounded by 0 < x < 1, 0 < z < x2 ? y2 . Ans: k=4/525Find k such that where V bounded by z = 0, z = x, x = 0, y = 0, andx + y = k. Ans: k=1What is the centroid of the cylinder V = {(x, y, z): x2 + y2 4, 0 z 1} in xyz-space with distances measured in feet if its density at (x, y, z) is cos z pounds per square foot?Ans: 16.5 Integrals in Cylindrical and Spherical CoordinatesCylindrical Coordinate SystemAny point (x, y, z) in R3 can be expressed in Cylindrical Coordinates as (r,θ,z) where r=x2+y2; tanθ=yx;z=z;x=rcosθ;y=rsin(θ) are the transformation equations, for 0≤r<∞, 0≤θ≤2π, -∞<z<∞.If c is a constant, r = c is a cylinder; θ=c is a plane; z = c is a plane.A triple integral in cylindrical coordinates is given by Rfx,y,zdVz1z2θ1θ2r1r2f(rcosθ, rsinθ, z) rdzdrdθ where the element of volume dV=rdzdrdθ .eg 36 Use a triple integral to find the volume bounded by the cylinder x2 + y2 = 4 and the sphere x2 + y2 + z2 = 9.50212945025In Cartesian Coordinates the volume is -22-4-x204-x2-9-x2-y29-x2-y2dzdydx. If we change to cylindrical coordinates, we have 02π02-9-r29-r2rdzdrdθ=4π327-55 cu units.eg 37 Compute the triple integral of f (x, y, z) = x2+y2 bounded by the paraboloid x2 + y2 + z = 1 and the plane z = 0.502129-1474-11-1-x201-x201-x2-y2x2+y2dzdydx=02π0101-r2r2dzdrdθ=02π01r21-r2drdθ=2π13-15=4π15.Spherical Coordinate SystemAny point (x, y, z) in R3 can be expressed in Spherical Coordinates as (ρ, θ,?) whereρ=x2+y2+z2; tanθ=yx; cos?=zρ=zx2+y2+z2; x=ρsin?cosθ;y=ρsin?sinθ;z=ρcos?(?) are the transformations equations for 0≤ρ<∞, 0≤θ≤2π, 0≤?≤πIf c is a constant, ρ=c is a sphere; θ=c is a plane; ?=c is a cone.9230277539700A triple integral in spherical coordinates is given by Rfx,y,zdV=?1?2θ1θ2ρ1ρ2f(ρsin(?)cosθ, ρsin(?)sinθ, ρcos(?))ρ2sin?dρdθd? where the element ofvolume dV=ρ2sin?dρdθd?.1174750228600eg 38 Find the volume bounded by the cone ?=π3 and the sphere ρ=4.V=0π302π04ρ2sin?dρdθd?=64π3 cu units. eg 39 Evaluate e(x2+y2+z2)32dV in the unit ball.V=0π02π01eρ3ρ2sin?dρdθd?=4π3 (e-1) .eg 40 Find the volume bounded by the cone z=x2+y2 and the sphere x2+y2+z2=z.122674878249Since ρ2=ρcos? from the sphere, 0<ρ<cos?, and ρcos?= ρsin(?)cosθ)2+(ρsin(?)sin(θ))2 =ρsin(?), we can say that the cone goes from, 0<?<π4. So =0π402π0cos?ρ2sin?dρdθd?=2π30π4(cos?)3sin?d?=π8 .Homework 16.5Describe the solid given by:a) 0≤θ≤π2, r≤z≤2; Ans: The first octant section of the cone x2+y2=z and the plane z = 2.b) 0≤ρ≤3, π2≤?≤π; Ans: The bottom half of the sphere of radius 3.c) 0≤?≤π4, ρ≤2; Ans: A snow coneFind the coordinate of the point ρ, ?, θ=(2, π4, π2) a) in the cylindrical (r, θ, z) coordinate systems. Ans: [(1, π2, 1)]b) in the Cartesian (x, y, z) coordinate systems. Ans: [(0, 1, 1)]Change to the other two coordinate system and sketcha) tan?=1. Ans: [x2+y2=z2; r2=z2 cone]b) r=2cos(θ). Ans: x2+y2=2x; ρsin?=2cosθ cylinder]Express ρsin?=2sin(θ) in rectangular coordinates and sketch the graph.Ans: [cylinder x2 + (y – 1)2 = 1]Express r=2sec(θ) in rectangular and spherical coordinates.Ans: [x=2; ρsin?cosθ=2]Express ρ=2sec(?) in rectangular and cylindrical coordinates. Ans: [z = 2]Describe the surface ρ=cos?. Find its distinctive points. (If the surface is a paraboloid, find its vertex; if a cylinder, find its radius; if a sphere, its center and radius etc.) Ans: sphere, r = ?, c:(0,0,1/2)Change cot?=1 to the cylindrical and Cartesian coordinates systems.Ans: z=r;z=x2+y2 inverted cone ]Calculate the volume of a sphere of radius R (a) by using cylindrical coordinates and (b) by using spherical coordinates. Ans: .Evaluate with V = {(x, y, z): 3 z }. Ans: 64.Use cylindrical coordinates to evaluate where V is the solid bounded by the cone and the plane z=1. Ans: 2π/3(a) is given in cylindrical coordinates. Rewrite the integral in rectangular coordinates. Ans: (a) Use spherical coordinates to evaluate where V is the quarter upper unit sphere centered at the origin with y ≥ 0 and z ≥ 0. Ans: π/15 is given in spherical coordinates. Express the integral in rectangular coordinates. Ans: ; The integral is in cylindrical coordinates. Express it as iterated integrals in spherical coordinates. Ans: ;16.7 Change of Variables in a Multiple Integral (Jacobians)Let f = f (x) and x = x (u). x=ax=bfxdx=u=a'u=b'f(xu)dxdudu where dxdu is the Jacobian of the transformation in R1.eg 41 If y=e2x+1 and u=2x+1. x=0x=1e2x+1dx=u=1u=3eudxdudu=, u=1u=3eu12du=e3-e2 where dxdu=12 is the called the Jacobian of the transformation.Transformation of Regions in R2A change of variables for double integrals is given by the transformation T from the u-v plane to the x-y plane where T (u, v) = (x, y) where x = g (u, v) and y = h (u, v). Let’s assume that T is a C1 transformation, this is that g and h have continuous first partial derivative. If T (ui, vi) = (xi, yi), the points (xi, yi) are called the image points of the points (ui, vi) under the transformation T. Let S be the region of all (ui, vi), if T transforms S into a region R in the x-y plane, R is called the image of S. If no two points in S have the same image, the transformation T is called one-to-one. If T is a one to one transformation, then it will have an inverse transformation T-1 that will transform points (xi, yi) into points (ui, vi), or (ui, vi) = T-1 (xi, yi).640152683eg 42 Draw in the x-y plane image of S : {(u, v)| 0 ≤ u ≤ 3, 0 ≤ v ≤ 2} under the transformationT = {x = 2u + 3v; y = u – v}.If we consider the positively oriented curve that enclose S, the line segments of the vertices of the rectangle are (0, 0), (3, 0), (3, 2), (0, 2) back to (0, 0). These four line segments in the u-v plane will map the region S to the region R in the x-y plane given by the line segments with vertices (0, 0), (6, 3), (12, 1), (6 – 1) back to (0, 0), with a cw orientation, under the transformation T.Since the transformation is Affine (linear), the lines segments of the rectangle will transform into line segments.eg 43 Draw in the x-y plane the image of S :{(r,θ)|0≤r≤2,0≤θ≤π2} under the non-Affine transformation T={x=rcosθ;y=rsin(θ)}.The line segments of the vertices of the rectangle of S are:r :0→2, θ=0 T (x :0→2, y=0) line segment(r=2, θ :0→π2) T x :2cosθ y=2sinθ quarter circle ccw θ :0→π2 (r :2→0, θ=π2) T x=0, y :2→0 line segment (r=0; θ : π2→0) T x=0, y=0 point eg 44 Find the image of S : {(u, v)| 0 ≤ u ≤ 1, 0 ≤ v ≤ 1} under the transformation the non-Affine transformation T = {x = u2 – v2; y = 2uv}.The line segments of the vertices of the rectangle of S are:u :0→1, v=0 T x=u2, y=0 line segment x :0→1 u=1, v :0→1 T (x=1-v2, y=2v) x=1-y24, x,y:1,0→(0,2) u :1→0, v=1 T x=u2-1, y=2ux=y24-1, x,y:0,2→(-1,0) u=0; v :1→0 T x=-v2, y=0 line segment x : -1→0 1623563131277Jacobians in R2 Let x = x (u, v) and y = y (u, v).y=cy=dx=ax=bfx,ydxdy=v=c'v=d'u=a'u=b'fxu,v,yu,v| Jxy|dudv whereJxy=?(x,y)?(u,v)=|yuxu yvxv| is the Jacobian of the transformation in R2 and |Jxy|dudv is the element of area of the transformation. If the transformation is Affine, the Jacobian is a constant.eg 45 Find the Jacobian in polar coordinatesSince x=rcosθ;y=rsinθ, Jxy=?(x,y)?(r,θ)=|yrxr yθxθ|=|sinθcosθ rcosθ-rsinθ|=r .The element of area dxdy in polar coordinates will be rdrdθ.Let x = x (u, v, w); y = y (u, v, w) and z = z (u, v, w) y=ey=fy=cy=dx=ax=bfx,ydxdy= w=e'w=f'v=c'v=d'u=a'u=b'fxu,v,w,yu,v,w, zu,v,w|Jxyz|dudvdw whereJxyz=?(x,y,z)?(u,v,w)=XuXvXwYuYvYwZuZvZw is the Jacobian of the transformation in R3.eg 46 Find the Jacobian in spherical coordinatesSince x=ρsin(?)cosθ;y=ρsin(?)sinθ;z=ρcos(?),Jxyz=?(x,y,z)?(ρ,θ,?)=XρXθX?YρYθY?ZρZθZ?=sin(?)cos(θ)-ρsin(?)sin(θ)ρcos(?)cos(θ)sin(?)sin(θ)ρsin(?)cos(θ)ρcos(?)sin(θ)cos(?)0-ρsin(?)=-ρ2sin(?) .The element of area dxdydz in spherical coordinates will be Jxyzdρdθd?=p2sin?dρdθd? Integration Under Transformation in R2eg 47 Evaluate ∫R x2+y2dA where R is the cylinder x2 + y2 = 1 using the transformation non-affine transformation T={x=rcosθ;y=rsinθ}.02π01r?(x,y)?(r,θ)drdθ. Since ?(x,y)?(r,θ)=xrxθyryθ=cosθ-rsinθsinθrcosθ=r,02π01r?(x,y)?(r,θ)drdθ=02π01r rdrdθ=2π3.eg 48 Evaluate ∫R x+ydxdy where R is the polygon with vertices (0, 0), (2, 3), (5, 1), (3, – 2) using the Affine transformation T = {x = 2u + 3v; y = 3u – 2v}.∫R x+ydxdy=∫S 5u+v?x,y?u,vdudv where ?(x,y)?(u,v)=xuxvyuyv=233-2=-13.To find the limits of integration, we need to find the region S under the transformation T. To do that, it is easier if we find the inverse transformation T-1.By Cramer’s Rule, T = {x = 2u + 3v; y = 3u – 2v} becomes T-1={u=2x+3y13;y=3x-2y13}.The line segments of the polygon (0, 0), (2, 3), (5, 1), (3, - 2) back to (0, 0) in the x-y are transformed to (0, 0), (1, 0), (1, 1), (0, 1) back to (0, 0) in the u-v. This is the square S : {(u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1}.So the integral becomes 01015u+v13 dudv=39.eg 49 Find Juv=?(u,v)?(x,y) for the previous problem.?(u,v)?(x,y)=uxuyvxvy=1132233-2=-113 . It can be shown that the Jacobians of Affine transformations and its inverse transformations are reciprocal of each other. So ?(x,y)?(u,v)=1?(u,v)?(x,y).eg 50 Evaluate ∫R x2+y2cosxydxdy where R is bounded by xy = 3, xy = – 3, x2 – y2 = 1 andx2 – y2 = 9.7868011102In this case the transformation is not given, but if we use the transformation T = {u = xy; v = x2 – y2}, the region of integration will be the rectangle – 3 ≤ u ≤ 3; 1 ≤ v ≤ 9.The integral ∫R x2+y2cosxydxdy=11-33x2+y2cos?(u)?(x,y)?(u,v)dudv. Since x and y are not expressed in terms of u and v we cannot find ?(x,y)?(u,v) directly.Since ?(x,y)?(u,v)=1?(u,v)?(x,y) also holds for non-Affine transformations, we can findJuv=uxuyvxvy=yx2x-2y=-2(x2+y2).The integral becomes 11-33x2+y2cos?(u)-12(x2+y2)dudv.=11-33cosu12dudv=8sin?(3) .eg 51 Evaluate ∫R x-yx+ydxdy where R is bounded by x – y = 0, x – y = 1, x + y = 1 and x + y = 3.1002030111125If we use the transformation T = {u = x – y; v = x + y}, the region of integration will be 0 ≤ u ≤ 1; 1 ≤ v ≤ 3.The integral ∫Rx-yx+ydxdy=1301uv?(x,y)?(u,v)dudv.Since ?(x,y)?(u,v)=1?(u,v)?(x,y), Juv=uxuyvxvy=1-111=2, so the integral becomes 1301uv12dudv=ln?(3)4 eg 52 Evaluate ∫R y dxdy where R is bounded by y2 = 4x + 4, y2 = – 4x + 4, y ≥ 0, under the transformation T = {x = u2 – v2; y = 2uv}.100246170389The condition y ≥ 0 implies that if y = 0, u = 0 or v = 0, and if y > 0, uv > 0 or uv < 0.Since y2 = 4x + 4 is the half parabola to the left of the region and y2 = – 4x + 4 is the half parabola to the right of the region, x and y into the left parabola gives(2uv)2=4u2-v2+4 or uv=u2-v2+1 with u=0?v=±1 v=0→no solution ,and x and y into the left parabola gives (2uv)2 = – 4 (u2 – v2) + 4 or uv = v2 – u2 + 1 withu=0→no solutionv=0→u=± 1 giving the region of integration 0 ≤ u ≤ 1; 0 ≤ v ≤ 1.The integral then becomes ∫R y dxdy=01012uv?(x,y)?(u,v)dudv=01012uv4u2+v2dudv=2, where ?(x,y)?(u,v)=4(u2+v2).Homework 16.7Draw in an xy-plane the image of the four squares in the figure below under the affine transformation x = 1 + 2u = v, y = 2 – u + 2v.Find an affine transformation T : x = a1u + b1v + c1, y = a2u + b2v + c2 that maps the square with corners = (0, 0), = (1, 0), = (0, 1) and = (1, 1) in a uv-plane into the parallelogram in the figure below. (b) What is the Jacobian of the transformation of part (a)? (c) Find the inverse to the transformation of part (a). (d) What is the Jacobian of the inverse transformation?Ans: The figures below show a region in a uv-plane and image R under the affine transformation x = u + v, y = u – v + 2. Use this transformation to convert into an integral with respect to u and v over. Ans: The figures below show a region in a uv-plane and its image R under the transformation x = 3/u, y = 4v1/3 – 3. Use this transformation to convert into an integral with respect to u and v over. Ans: Figure 5Figure 6What is the Jacobian of the affine transformation x = u +2v + 3w, y = 4u – 5v, z = 6v + 2w? (b) Suppose that a solid in uvw-space has volume 100 cubic meters. What is the volume of its inverse under the transformation of part (a)? 46; 4600m3.Draw in an xy-plane the image under the affine transformation x = 2u + 5v, y = –2u + 2v of the pentagon in the figure below. Find the values of the integrals by making affine changes of variables to obtain integrals over boxes with sides parallel to the coordinate planes: , with V = {(x, y, z): 0 x + y – z 2, 0 x – y + z 3, 0 y + 2z 4 Ans; Arctan(4)Find the values of the integrals by making affine changes of variables to obtain integrals over boxes with sides parallel to the coordinate planes: , with V = {(x, y, z): 1 x + y 2, 3 x – y 4, 4 x + y + z 5. Ans: 21/8 ................
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