# 425notes9 - Math 425 Notes 9 Multiple Integrals Dention We...

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Math 425 Notes 9Multiple Integrals: DefintionWe state a rough version of the fundamental theorem of calculus. Almost all calcula-tions involving integrals lead back to this.Definition 1.Letf:RRand let [a, b] be a closed interval. Leta=x0< x1<· · ·< xn=bbe a division of the interval [a, b].For eachiletx*i[xi, xi-1].A sum of the formi=nXi=1f(x*i)(xi-xi-1)is called a Riemann sum.Notation:Keep the notation from the definition above.Zbaf(x)dx=limi=nXi=1f(x*i)(xi-xi-1)Here the limit is taken over partitions as the length of the pieces all go to zero.Theorem1.Fundamental Theorem of CalculusLetf:RRbe continuous. ThenZbaf(x)dx=F(a)-F(b)wheredFdx=f.Notation: Letxi, xi-1R, then Δxi=xi-xi-1.Definition 2.LetRbe the rectangle [a, b]×[c, d]R2.Letf:RRbe a function onR.Letxi=a+in(b-a), yj=c+jn(d-c).ThenZ ZRfdxdy= limn→∞i,j=nXi,j=1f(xi, yjxiΔyjprovided the limit exists.Theorem2.Assume thatfis continuous on a rectangleRexcept on points in the unionof a finite number of graphs. Then the limit in the definition above exists.1
Assume thatf(x, y) =zis the graph of a continuous (except possibly on points in thefinite union of graphs) non-negative function. ThenR RRfdxdyis the volume of the solidabove thez= 0 plane and below the graph off.Here is a fancier case. How do we find the volume under the functionf(x, y), abovethe planez= 0 and within the triangle with vertices (0,0),(1,0),(1,1).Apparently this isnot covered by the above theorems. We extend the functionfby making it equal to zerooutside the triangle and using the rectangle [0,1]×[0,1].The singularities are all on theliney=xand hence are allowable.How do we compute these double integrals.These integrals are the limits of sumsin a rectangular array.Here are two ways of organizing the summing of numbers in arectangular array.The first method is to add the numbers in each column and thenadding these sums. A second method is to add the numbers in each row and then addingthese sums.Formally we havei,j=nXi,j=1f(xi, yjxiΔyj=j=nXj=1i=nXi=1f(xi, yjxi!Δyj=i=nXi=1j=nXj=1f(xi, yjyj!Δxi.If we take the limit asn→ ∞, then these three sums have the same limit providedfis bounded, and continuous outside a finite union of graphs. This is a delicate result.Theorem3.Assume thatfis nice on a rectangleR= [a, b]×[c, d].The first sumconverges toZ ZRfdxdy.The second converges toZdc(Zbafdx)dy.The third converges toZba(Zdcfdx)dy.These are all equal.Example4.LetR= [-1,2]×[3,4], f(x, y) = 2x2+xy+y2.ThenZ ZR=Z43(Z2-1f(x, y)dx)dy.We separate out the inside integral to getZ2-1(2x2+xy+y2)dx=x3+x2y+xy2|x=2x=-1= 3y2+ (3/2)y+ 9.
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