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.Chapter 14 Multiple Integrals.1. Double Integrals, Iterated Integrals, Cross-sections2. Double Integrals over more general regions, Definition,Evaluation of Double Integrals, Properties of Double Integrals. Area and Volume by Double Integration, Volume by IteratedIntegrals, Volume between Two surfaces.4 Double Integrals in Polar Coordinates, More general Regions3. Applications of Double Integrals, Volume and First Theorem ofPappus, Surface Area and Second Theorem of Pappus,Moments of Inertia6. Triple Integrals, Iterated Triple Integrals5. Integration in Cylindrical and Spherical Coordinates8. Surface Area, Surface Area of Parametric Surfaces, Surfaces7.Area in Cylindrical Coordinates.9 Change of Variables in Multiple Integrals, Jacobian.Math 200 in 2011.

.Chapter 14 Multiple Integrals.14.2 Properties of double integrals14.3 Area and volume of double integration14.4 Double integrals and iterated integral in polar coordinates14.4 Gaussian probability distribution.Math 200 in 2011.

. 1. D2D( f (x, y) g(x, y) )dA cf (x, y) dA cD Df (x, y) dA Dg(x, y) dA.f (x, y) dA. If f (x, y) g(x, y) for all (x, y) D, then f (x, y) dA g(x, y) dA.34. DD Df (x, y) dA D1 f (x, y) dA D2f (x, y) dA, whereD D1 D2 , and D1 and D2 donot overlap except perhaps on theirboundary. 5DdA D1 dA Area of D A(D). If m f (x, y) M for all (x, y) D, then 6mA(D) .Df (x, y) dA MA(D).Math 200 in 2011.

. Double Integrals in Polar Coordinates.Proposition. If f (x, y) is continuous on a region D in xy-plane, and Dcan be described in polar coordinates in the following form{ (r, θ ) α θ β, g1 (θ ) r g2 (θ ) }, then .Df (x, y) dA β g2 (θ )αg1 ( θ )f (r cos θ, r sin θ ) r dr dθ.Math 200 in 2011.

Given a region R in xy-plane, sometimes it is rather easy to describethe boundary of R in terms of polar coordinates instead of rectangularcoordinates. An example is x2 y2 a2 , can be easily described as{ (r, θ ) 0 θ 2π, 0 r a }. It is the reason why one needs todevelop the double integral in terms of polar coordinates, just like theone in rectangular coordinates.Without really getting into the details, one can subdivide the planeregion R in terms of polar parameters, just like defining doubleintegral, so that we have a sectorial area A r r θ, so theRiemann sum is written f (ri cos θj , ri sin θj ) ri r θ, which willijconverges to the double integral R′f (r cos θ, r sin θ ) r dr dθ, where R′is the domain for θ and r representing the domain R in xy-plane.Math 200 in 2011.

.Example. Use a double integral to find thearea enclosed by one loop of thethree-leavedrose r sin 3θ.Solution. The region, in polar coordinates, is{ (r, θ ) 0 θ π/3, 0 r sin 3θ },hence the area of the region D is π/3 sin 3θ π/3 1 π/312r dr dθ sin (3θ )dθ (1 cos(6θ ) ) dθ 24 000[]π/3 011πθ sin(6θ ) .46120.Math 200 in 2011.

.2 1, y 0 }Example. D { (x, y) x2 y 22. { (x, y) 0 y 1, 1 y x 1 y }.One can describe D in terms of polar coordinates as follows{ (r, θ ) π2 θ π2 , 0 r 1 }.Example. For any a 0, prove that the equation r 2a sin θ in polar22.coordinates is the circle x y 2ay.Solution. The curve x2 y2 2ax can be described in polarcoordinates as r2 r2 (cos2 θ sin2 θ ) x2 y2 2ay 2ar sin θ, soone has r 2a sin θ (0 θ π ). Geometrically, it means aright-angled triangle circumscribed in a circle of diameter 2a.Math 200 in 2011.

.Example. Find the volume of the solid D bounded by the paraboloid22S. : z 25 x y and the xy-plane.Solution. The paraboloid S : z 25 x2 y2intersect the xy-plane π : z 0 in the curve C :0 25 x2 y2 , which is a circle x2 y2 52 .So the shadow R of the solid D after projectingonto xy-plane is given by the circular disc R { (x, y) x2 y2 52 }, in polar coordinates isgiven by { (r, θ ) 0 θ 2π, 0 r 5}.Then the volume of the solid D is given by R(25 x2 y2 )dA 52π0 2π 5(25r r3 ) dr 2π0(25 r2 ) r dr dθ ]5r4625 π.24 020[25r2.Math 200 in 2011.

.Example. Find the volume of the solid that lies under the paraboloidz. x2 y2 , above the xy-plane, and inside the cylinder x2 y2 2x.Solution. The cylinder x2 y2 2x lies overthe circular disk D which can be described as{ (r, θ ) π/2 θ π/2, 0 r 2r cos θ }in polar coordinates. The reason is that if wewrite (x, y, z) (r cos θ, r sin θ, z) for any pointin the cylinder, then r2 x2 y2 2x 2r cos θ, i.e. r 2 cos θ. As 2 cos θ r 0,it follows that π/2 θ π/2. The height of the solid is the z-value of theparaboloid from the xy-plane. Hence the volume V of the solid is π/2 2 cos θ3π(x2 y2 ) dA r2 · r dr dθ .2D π/2 0.Math 200 in 2011.

.Example.Find the volume of the solid that lies below the hemisphere z 9 x2 y2 , above the xy-plane, and inside the cylinder22.x y 1.Solution. Let R be the shadow of D after projecting on xy-plane, then R is the circular disk centered at the origin with radius 1, in polar coordinates { (r, θ ) 0 r 1, 0 θ 2π }. Moreover, the top ztop 9 x2 y2 9 r2 , andzbottom 0.Hencevolume of the solid D is 2π the 1 2π 1 29 r · r dr dθ 9 r2 d(8 r2 ) 2 00]1[0] 2π [ 3/22π(9 r2 )3/2 9 83/2 (27 16 2). π3/2330.Math 200 in 2011.

.Example. Find the volume of the solid bounded above by theparaboloidz 8 x2 y2 , and below by the paraboloid z x2 y2 .Solution. Let P(x, y, z) be the intersection of twoparaboloids, then one has 8 x2 y2 z x2 y2 , so x2 y2 4 22 , which is a circle. Theshadow R of the solid D is then the circular disc,in polar coordinates { (r, θ ) 0 r 2, 0 θ 2π }. As 0 r 2, we have r2 4, i.e. r2 8 r2 ,so one knows that the top of the solid is given byztop 8 x2 y2 8 r2 , and the bottom of thesolid is given by zbottom x2 y2 r2 .Hencevolume of the solid D is 2π the22(8 r2 r2 ) · r dr dθ 2π(8r r3 ) dr000]2[r4 24π. 2π 4r2 4 0.Math 200 in 2011.

.Example. Determine the volume of the region D common to theinteriors of the cylinders x2 y2 1 and x2 z2 1, in the firstoctant.Solution. One immediately recognizes the solid D has a top given byx2 z2 1, i.e. zmax (x) 1 x2 , and xy-plane as the bottom.Moreover, the shadow R of the solid D is a circle disk in the 1stquadrant, so R { (x, y) 0 x 1, 0 y 1 x2 }. The volume of D is given by( 1 x2 0) dAR[]1 1 1 1 x2 x32221 x dy dx (1 x )dx x .330000.Math 200 in 2011.

.Example. Determine the volume of the region D common to theinteriors of the cylinders x2 y2 1 and x2 z2 1, in the firstoctant.Solution. One immediately recognizes the solid D has a top given byx2 z2 1, i.e. zmax (x) 1 x2 , and xy-plane as the bottom.Moreover, the shadow R of the solid D is a circle disk in the 1stquadrant, so R can be described in terms of polar coordinates as{ (r, θ ) 0 r 1, 0 θ π/2 }. The volume of D is given by π/2 1 1 r2 cos2 θ rdrdθ 23 .00(*): the answer 32 is given by integration via rectangular coordinates.Math 200 in 2011.

.Example. Show that the volume of the solid region E bounded by thethreecylinders x2 y2 1, y2 z2 1 and x2 z2 1 is 16 8 2.Solution. The solid is symmetric with respect to the 3 coordinate axes. Thus, it suffices to compute the volume of the portion Dwith nonnegative x, y, z-coordinates. In otherwords, D lies in the first octant of the coordinate system.Math 200 in 2011.

.Example. Show that the volume of the solid region E bounded by thethreecylinders x2 y2 1, y2 z2 1 and x2 z2 1 is 16 8 2.Solution. The solid is symmetric with respect to the 3 coordinate axes. Thus, it suffices to compute the volume of the portion Dwith nonnegative x, y, z-coordinates. In otherwords, D lies in the first octant of the coordinate system. 2For the solid D, it is bounded on top by the graphs of z 1 x andz 1 y2 . On its side and bottom, it is bounded by the cylinderx2 y2 1 and the three coordinate planes. Furthermore, the graphsof z 1 x2 and z 1 y2 intersect along a curve on the planex y. Thus the solid is under the graph of z 1 x2 over theregion D in xy-plane described as { (r, θ ) 0 r 1, 0 θ π/4 }in the polarvolume of original solid E coordinates. Hence, π/4 the1 16 D 1 x2 dA 16 01 r2 cos2 θrdrdθ 0][r 1 π/4 1 sin3 θ π/42 cos2 θ )3/2dθ 16 0dθ 16 0 (1 r cosθ3 cos2 θr 0 π/416 16 8 2.3 [tan θ sec θ cos θ ]0.Math 200 in 2011.

. π.20 M()Solution. Ase x dx lime x dx lim 1 e M 1,Example. Prove with the polar-coordinates that.M 00e x2dx M it follows from comparison test with 0 e x e x that the improper 2 Me x dx converges. Let IM e x dx. Need to prove00 ππ2that lim IM . It suffices to show that lim IM.24M M integral2By Fubini( theorem,)we(have M2IM RMee x dx ·0 x2 y22 M0e y dy2)2 M M 00e x e y dy dx 22dA, where RM [0, M] [0, M] is a square of length M.Continue.Math 200 in 2011.

.Example. Prove with the polar-coordinates that. 0e x2 dx π.2Solution. Let DM be quarter circular disk of radius M centered atorigin. By means of polar coordinates, we have JM 2 y2DM e xdA π/2 M00e r r dr dθ22π M r2 1π e d( r2 ) (1 e M ).2 024πIt follows that lim JM .4M 22As f (x, y) e x y 0, and DM RM DM 2 , it follows from2 J . The result followsproperty of double integral that JM IMM 2from sandwich theorem thatππ2 lim JM lim IM lim JM 2 lim JM .44M M M M .Math 200 in 2011.

.Applications of Double Integrals.Suppose a (planar) object, in region R, is made of different material inwhich the density (mass per unit area) is given by δ(x, y), dependingon the location (x, y). Then the total mass of R is given approximatelyby the Riemann sum i δ(xi , yi ) Ai , which will converges to thedouble integral m .Dδ(x, y) dA. We call it the mass of the object.Similarly,(one can define the center of mass (centroid)of the object by) 11(x, y) xδ(x, y) dA,yδ(x, y) dA .m Dm D.Math 200 in 2011.

.Example. A lamina R is made of the part of circular disc of radius a in1st quadrant. Its density is proportional to the distance from the.origin. Determine the position of its centroid.Solution. In terms of polar coordinates, D can be describedas { (r, θ ) 0 r a, 0 θ π/2 }. Let δ(x, y) k x2 y2 kr. π/2 aπ ka3kπa3Then the mass m kr · r dr dθ · .2 3600As all the conditions on the lamina R are symmetric in x and y, so 1 π/2 ax y (r sin θ ) · kr · r dr dθ ( π/2m 0)0 ( a)kka43a3sin θdθ ·r dr 1 .3m42πkπa/600.Math 200 in 2011.

.First Theorem of Pappus: Volume of Revolution.Suppose that a plane region R is revolved around an axis in its planegenerating a solid of revolution with volume V. Assume that the axisdoes not intersect the interior of R. Then the volume V of the solid isV A · d,where A is the area of R and d is the distance traveled by the centroid.of R.Remark. Cut the region R into vertical strips, and each vertical stripafter rotating will form a ring, which contributes Vi 2πxi Ai ,where Ai is the area of the vertical strip. It follows from Riemann sumthat.nnMath 200 in 2011.

.Volume of a Sphere.Provethat the volume of a sphere is 43 πa3 .Solution. Let R be a region bounded by the upper semi-circle x2 y2 a2 , y 0 and x-axis.In polar coordinates, R can be described as{ (r, θ ) 0 θ π, 0 r a }. Using polar coordinates, the area of π aa2r dr dθ R is π πa2 /2. Similarly, y of R is given by200 π a11ydA r sin θ · r dr dθπa2 /2 R(πa2 /2) ( 0 a 0 )[ 3 ]a π11r2π sinθdθrdr [ cosθ] 03 0πa2 /2πa2 /2004a2a3 . By Pappus’s theorem, the volume of a sphere 2 33ππa24a πa24of radius a 2πy Area of R 2π ·· πa3 .3π23.Math 200 in 2011.

.Second Theorem of Pappus: Surface Area of Revolution.Suppose that a plane curve C is revolved around an axis in its planethat does not intersect the curve. Then the area A of the surface ofrevolution generated isA s · d,where s is the arc-length of C and d is the distance traveled by the.centroid of C.Math 200 in 2011.

.Let R be a plane lamina and ℓ be a straight line that may or may notlie in xy-plane. Then the moment of inertia I of R around the axis ℓ isRp2 δ(x, y)dA, where p p(x, y) is the shortest distance from thepoint (x, y) of R to the line ℓ, and δ(x, y) is the density of R at the point.(x, y).For the coordinate axii, we have. Ix Ix-axis Iz Iz-axis . RR(y2 )δ(x, y)dA, Iy Iy-axis R(x2 )δ(x, y)dA and(x2 y2 )δ(x, y)dA.For any plane laminaR, Define the center (x̂, ŷ, ẑ) of gyration by IyIxIzx̂ m , ŷ m , ẑ m.Math 200 in 2011.

Given a region R in xy-plane, sometimes it is rather easy to describe the boundary of R in terms of polar coordinates instead of rectangular coordinates. An example is x2 y2 a2, can be easily described as {(r,q) 0 q 2p, 0 r a }.It is the reason why one needs to develop the double integral in t